MOTION IN A PLANE-Exercise
Physics - Exercise
Q1. State, for each of the following physical quantities, if it is a scalar or a vector :
volume, mass, speed, acceleration, density, number of moles, velocity, angular
frequency, displacement, angular velocity.
Solution
| Physical Quantity | Type |
|---|---|
| Volume | Scalar |
| Mass | Scalar |
| Speed | Scalar |
| Acceleration | Vector |
| Density | Scalar |
| Number of moles | Scalar |
| Velocity | Vector |
| Angular frequency | Scalar |
| Displacement | Vector |
| Angular velocity | Vector |
Q2. Pick out the two scalar quantities in the following list :
force, angular momentum, work, current, linear momentum, electric field, average
velocity, magnetic moment, relative velocity.
Solution
The two scalar quantities in the given list are: work and current.
Work has only magnitude and is defined as the dot product of force and displacement, so it is a scalar quantity.
Electric current is treated as a scalar because it follows ordinary algebraic addition rather than vector addition laws
Q3. Pick out the only vector quantity in the following list :
Temperature, pressure, impulse, time, power, total path length, energy, gravitational
potential, coefficient of friction, charge.
Solution
The only vector quantity in the given list is impulse.
Impulse has both magnitude and direction because it equals the change in momentum, and momentum is a vector quantity.
Q4. State with reasons, whether the following algebraic operations with scalar and vector
physical quantities are meaningful :
(a) adding any two scalars,
(b) adding a scalar to a vector of the same dimensions ,
(c) multiplying any vector by any scalar,
(d) multiplying any two scalars,
(e) adding any two vectors,
(f) adding a component of a vector to the same vector.
Solution
(a) Adding any two scalars – Meaningful only if they represent the same physical quantity (same unit), e.g. adding two masses or two times; otherwise physically meaningless.
(b) Adding a scalar to a vector of the same dimensions – Not meaningful, because addition is defined only scalar–scalar or vector–vector, never scalar–vector.
(c) Multiplying any vector by any scalar – Meaningful, as the result is a vector whose magnitude is scaled by the scalar while direction remains unchanged.
(d) Multiplying any two scalars – Meaningful, since the product is another scalar (e.g. density × volume = mass).
(e) Adding any two vectors – Meaningful only if they are of the same physical quantity and unit (e.g. two forces, two displacements); otherwise not physically meaningful.
(f) Adding a component of a vector to the same vector – Meaningful, because the component is itself a vector of the same kind, so this is just ordinary vector addition.
Q5. Read each statement below carefully and state with reasons, if it is true or false :
(a) The magnitude of a vector is always a scalar,
(b) each component of a vector is always a scalar,
(c) the total path length is always equal to the magnitude of the
displacement vector of a particle.
(d) the average speed of a particle (defined as total
path length divided by the time taken to cover the path) is either greater or equal to
the magnitude of average velocity of the particle over the same interval of time,
(e) Three vectors not lying in a plane can never add up to give a null vector.
Solution
(a) The magnitude of a vector is always a scalar – True.
The magnitude is just a numerical measure (with units) and has no direction, so it is a scalar.
(b) Each component of a vector is always a scalar – False.
Each component (like \(A_x,\ A_y,\ A_z\)) is itself a vector along its coordinate axis, not a scalar.
(c) The total path length is always equal to the magnitude of the displacement vector – False.
Total path length is a scalar and is generally greater than or equal to the magnitude of displacement; they
are
equal only for straight-line motion in one direction.
(d) The average speed is either greater or equal to the magnitude of average velocity – True.
Average speed = total path length / time, while magnitude of average velocity = displacement / time, and
path
length is always ≥ displacement.
(e) Three vectors not lying in a plane can never add up to give a null vector – True.
For three vectors to sum to the zero vector, they must form the sides of a triangle taken in order, which is
only possible if all three lie in the same plane.
Q6. Establish the following vector inequalities geometrically or otherwise :
(a) |a+b| < |a+b|
(b) |a+b|> ||a| −|b||
(c) |a−b| < |a| + |b|
(d) |a−b|> ||a| −− |b||
When does the equality sign above apply?
Solution
The required inequalities are the standard triangle inequalities for vectors, and equality holds only in the collinear cases described below.
Inequalities for \(\mathbf{a}+\mathbf{b}\)
\[\mid\mathbf{a}+\mathbf{b}\mid\le\mid\mathbf{a}\mid+\mid\mathbf{b}\mid\]This follows from the triangle inequality: in the triangle formed by \(\mathbf{a},\mathbf{b}\) and \(\mathbf{a}+\mathbf{b}\), the side corresponding to \(\mathbf{a}+\mathbf{b}\) cannot exceed the sum of the other two sides.
\[\mid\mathbf{a}+\mathbf{b}\mid\geq\mid\mid\mathbf{a}\mid-\mid\mathbf{b}\mid\mid\]Geometrically, the length of any side of a triangle is at least the difference of the lengths of the other two sides, giving this lower bound for \(\mid\mathbf{a}+\mathbf{b}\mid\). Equality:
\(\mid\mathbf{a}+\mathbf{b}\mid=\mid\mathbf{a}\mid+\mid\mathbf{b}\mid\) when \(\mathbf{a}\) and \(\mathbf{b}\) are collinear and in the same direction \((\angle 0^\circ)\).
\(\mid\mathbf{a}+\mathbf{b}\mid=\mid\mid\mathbf{a}\mid-\mid\mathbf{b}\mid\mid\) when \(\mathbf{a}\) and \(\mathbf{b}\) are collinear and in opposite directions \((\angle {180}^\circ)\).
Inequalities for \(\mathbf{a}-\mathbf{b}\)
Viewing \(\mathbf{a}-\mathbf{b}=\mathbf{a}+(-\mathbf{b})\), the same triangle inequalities apply with \(-\mathbf{b}\).
\[\mid\mathbf{a}-\mathbf{b}\mid\le\mid\mathbf{a}\mid+\mid\mathbf{b}\mid\]This is again the triangle inequality, now for \(\mathbf{a}\) and \(-\mathbf{b}\).
\[\mid\mathbf{a}-\mathbf{b}\mid\geq\mid\mid\mathbf{a}\mid-\mid\mathbf{b}\mid\mid\]Here \(\mid\mathbf{a}-\mathbf{b}\mid\) cannot be smaller than the difference of the magnitudes of \(\mathbf{a}\) and \(\mathbf{b}\).
Equality:
\(\mid\mathbf{a}-\mathbf{b}\mid=\mid\mathbf{a}\mid+\mid\mathbf{b}\mid\) when \(\mathbf{a}\) and \(\mathbf{b}\) are collinear and in opposite directions (so \(-\mathbf{b}\) is in the same direction as \(\mathbf{a}\)).
\(\mid\mathbf{a}-\mathbf{b}\mid=\mid\mid\mathbf{a}\mid-\mid\mathbf{b}\mid\mid\) when \(\mathbf{a}\) and \(\mathbf{b}\) are collinear and in the same direction.
Q7. Given a + b + c + d = 0, which of the following
statements are correct :
(a) a, b, c, and d must each be a null vector,
(b) The magnitude of (a + c) equals the magnitude of ( b +d),
(c) The magnitude of a can never be greater than the
sum of the magnitudes of b, c, and d,
(d) b + c must lie in the plane of a and d if a and d are
not collinear, and in the line of a and d, if they are
collinear ?
Solution
From \(\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=0\), the correct statements are (b), (c) and (d).
Statement (a)
(a) \(\mathbf{a},\ \mathbf{b},\ \mathbf{c},\ \mathbf{d}\) must each be a null vector – False.
Non-zero vectors can sum to zero by forming a closed polygon, so each vector need not be zero.
Statement (b)
(b) The magnitude of \mathbf{a}+\mathbf{c} equals the magnitude of \mathbf{b}+\mathbf{d} – True.
From \(\mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=0,\
\mathbf{a}+\mathbf{c}=-(\mathbf{b}+\mathbf{d})\), so these two vectors are equal in magnitude and opposite
in
direction.
Statement (c)
(c) The magnitude of \(\mathbf{a}\) can never be greater than the sum of the magnitudes of
\(\mathbf{b},\ \mathbf{c},\ \mathbf{d}\) – True.
Rewriting, \(\mathbf{a}=-(\mathbf{b}+\mathbf{c}+\mathbf{d})\), hence
\[\mid\mathbf{a}\mid=\mid\mathbf{b}+\mathbf{c}+\mathbf{d}\mid\le\mid\mathbf{b}\mid+\mid\mathbf{c}\mid+\mid\mathbf{d}\mid\]
by the triangle inequality.
Statement (d)
(d) \(\mathbf{b}+\mathbf{c}\) must lie in the plane of \(\mathbf{a}\) and \(\mathbf{d}\) if \(\mathbf{a}\)
and \(\mathbf{d}\) are
not collinear, and in the line of \(\mathbf{a}\) and \(\mathbf{d}\), if they are collinear – True.
Since \[\mathbf{a}+(\mathbf{b}+\mathbf{c})+\mathbf{d}=0\] the three vectors
\(\mathbf{a},\mathbf{d},(\mathbf{b}+\mathbf{c})\) must form a triangle; thus \[\mathbf{b}+\mathbf{c}\] lies
in the plane
defined by \(\mathbf{a}\) and \(\mathbf{d}\), or on their line if they are collinear.
Q8. Three girls skating on a circular ice ground of radius 200 m start from a point P on the edge of the ground and reach a point Q diametrically opposite to P following different paths as shown in Fig. 3.19. What is the magnitude of the displacement vector for each ? For which girl is this equal to the actual length of path skate ?
Solution
Each girl has the same displacement magnitude of 400 m, and this equals the actual path length for the girl who skates straight along the diameter.
Displacement of each girl
- Initial point is P and final point is Q, which are diametrically opposite on a circle of radius 200 m.
- Displacement is the straight-line distance PQ = diameter =2×200=400 m for each girl, regardless of the curved path taken.
Girl for whom displacement = path length
Only the girl who skates directly along the diameter PQ has her distance travelled equal to 400 m, so for her the displacement equals the actual length of the path.
Solution
Given: Radius of the circular park \( r = 1 \text{ km} \) and
total time of the round trip \[\begin{aligned}t &=
10 \text{ min} \\\\&= \dfrac{10}{60} \text{ h} \\\\&= \dfrac{1}{6} \text{ h} \end{aligned}\]
The cyclist starts from the centre \( O \), moves to the point \( P \) on the circumference, then follows the circular arc from \( P \) to \( Q \), and finally returns from \( Q \) to \( O \) along the radius. The segments \( OP \) and \( QO \) are radii of the circle, and the arc \( PQ \) subtends an angle of \( 90^{\circ} \) at the centre, so it is one–fourth of the full circumference.
The total path length is \[\begin{aligned} s &= OP + \text{arc } PQ + QO\\\\ &= r + \frac{2\pi r}{4} + r\\\\ &= 2r + \frac{\pi r}{2} \end{aligned}\]
Substituting \( r = 1 \text{ km} \),
\[\begin{aligned} s &= 2 + \frac{\pi}{2}\\\\ &\approx 2 + \frac{3.14}{2}\\\\ &= 2 + 1.57\\\\ &= 3.57 \text{ km} \end{aligned}\]Since the cyclist starts from \( O \) and returns to the same point, the initial and final positions are identical, so the net displacement vector is zero. Thus, \[ \text{Net displacement} = 0 \]
Average velocity is defined as displacement divided by the total time taken. Because the net displacement is zero, \[\begin{aligned} \bar{v} &= \frac{\text{displacement}}{t} \\\\&= \frac{0}{1/6} \\\\&= 0 \end{aligned}\]
Average speed is defined as total distance travelled divided by total time taken. Therefore, \[\begin{aligned} v_{\text{avg}} &= \frac{s}{t}\\\\ &= \frac{3.57}{1/6}\\\\ &= 3.57 \times 6\\\\ &= 21.42 \text{ km h}^{-1} \end{aligned}\]
Q10. On an open ground, a motorist follows a track that turns to his left by an angle of 600 after every 500 m. Starting from a given turn, specify the displacement of the motorist at the third, sixth and eighth turn. Compare the magnitude of the displacement with the total path length covered by the motorist in each case.
Solution
A motorist travels 500 m along a straight segment and then turns left by \(60^{\circ}\) after each segment. Thus, each side of the track is a straight line of length \(500 \text{ m}\), and successive sides make an internal angle of \(60^{\circ}\) with each other.
Third turn: After three such equal turns, the motorist has traced three sides of a regular hexagon. The total path length up to the third turn is \[\begin{aligned} \text{Total distance} &= 3 \times 500 \\&= 1500 \text{ m}. \end{aligned}\] The displacement is the straight-line distance between the starting point and the position after the third turn. Geometrically, this is the third side of an equilateral triangle formed by the first and the third sides, giving \[ \text{Displacement} = 1000 \text{ m} \] (from the detailed vector addition or geometry, using components of \(500 \text{ m}\) at \(60^{\circ}\) intervals). The displacement \(1000 \text{ m}\) is less than the total path length \(1500 \text{ m}\).
Sixth turn: After six turns, the motorist completes a full regular hexagon and returns to the initial point. The total path length is \[\begin{aligned} \text{Total distance} &= 6 \times 500 \\&= 3000 \text{ m}. \end{aligned}\] Since the starting and final points coincide, the displacement is \[ \text{Displacement} = 0. \] Here the displacement is zero, while the total distance covered is \(3000 \text{ m}\).
Eighth turn: After six turns the motorist is back at the starting point and then continues for two more sides of 500 m each. Therefore, the total path length up to the eighth turn is \[\begin{aligned} \text{Total distance} &= 3000 + 500 + 500 \\&= 4000 \text{ m}. \end{aligned}\] The net displacement after eight turns corresponds to the vector sum of two sides of the hexagon separated by \(60^{\circ}\). Using vector components, \[\begin{aligned} \text{Displacement} &= 2 \times 500 \cos 30^{\circ}\\\\ &= 2 \times 500 \times \frac{\sqrt{3}}{2}\\\\ &= 500\sqrt{3}\\\\ &\approx 866 \text{ m}. \end{aligned}\] Thus, in this case the displacement is about \(866 \text{ m}\), which is much smaller than the total distance \(4000 \text{ m}\).
Q11. A passenger arriving in a new town wishes to go from the station to a hotel located 10 km away on a straight road from the station. A dishonest cabman takes him along a circuitous path 23 km long and reaches the hotel in 28 min. What is (a) the average speed of the taxi, (b) the magnitude of average velocity ? Are the two equal ?
Solution
A passenger has to travel from the station to a hotel that is \(10 \text{ km}\) away along a straight road. The cabman, however, follows a circuitous route of total length \(23 \text{ km}\) and takes \(28 \text{ min}\) to reach the hotel.
Total distance travelled by the taxi is \[ \text{Distance} = 23 \text{ km}. \] The time taken is \[ t = 28 \text{ min} = \frac{28}{60} \text{ h}. \] The displacement is simply the straight-line distance from the station to the hotel, which is \[ \text{Displacement} = 10 \text{ km}. \]
The average speed of the taxi is defined as total distance travelled divided by the total time taken. Hence, \[ v_{\text{avg}} = \frac{23}{28/60} = \frac{23 \times 60}{28} \approx 49.3 \text{ km h}^{-1}. \]
The magnitude of the average velocity is defined as the magnitude of displacement divided by the total time taken. Therefore, \[ |\vec{v}_{\text{avg}}| = \frac{10}{28/60} = \frac{10 \times 60}{28} \approx 21.4 \text{ km h}^{-1}. \]
Thus, the average speed of the taxi is about \(49.3 \text{ km h}^{-1}\) while the magnitude of the average velocity is about \(21.4 \text{ km h}^{-1}\). These are not equal because the path taken is not a straight line and so the distance travelled is greater than the magnitude of the displacement.
Q12. The ceiling of a long hall is 25 m high. What is the maximum horizontal distance that a ball thrown with a speed of 40 m s-1 can go without hitting the ceiling of the hall ?
Solution
The ceiling of the hall is at a height \( h = 25 \text{ m} \) and
the ball is thrown with speed \( u = 40 \text{ m s}^{-1} \).
For the ball to just avoid hitting the ceiling, its maximum height must be equal to
\(25 \text{ m}\).
For a projectile, the maximum height is given by
\[
h = \frac{u^{2} \sin^{2}\theta}{2g},
\]
where \( \theta \) is the angle of projection and \( g \approx 10 \text{ m s}^{-2} \)
Substituting the values, \[\begin{aligned} 25 &= \frac{40^{2} \sin^{2}\theta}{2 \times 10}\\\\ \Rightarrow 25 \times 2 \times 10 &= 40^{2} \sin^{2}\theta\\\\ \Rightarrow \sin^{2}\theta &= \frac{25 \times 2 \times 10}{40 \times 40}\\\\ &= \frac{5}{16} \end{aligned}\] Hence, \[ \sin\theta = \frac{\sqrt{5}}{4} \]
Using \( \sin^{2}\theta + \cos^{2}\theta = 1 \), \[\begin{aligned} \cos^{2}\theta &= 1 - \frac{5}{16}\\\\ &= \frac{16 - 5}{16}\\\\ &= \frac{11}{16}, \end{aligned}\] so \[ \cos\theta = \frac{\sqrt{11}}{4}. \]
The horizontal range of the projectile is \[\begin{aligned} R &= \frac{u^{2} \sin 2\theta}{g}\\\\ &= \frac{u^{2} (2 \sin\theta \cos\theta)}{g} \end{aligned}\] Substituting the known values, \[\begin{aligned} R &= \frac{40 \times 40 \times 2 \sin\theta \cos\theta}{10}\\\\ &= \frac{40 \times 40 \times 2 \times \frac{\sqrt{5}}{4} \times \frac{\sqrt{11}}{4}}{10}\\\\ &= \frac{40 \times 40 \times 2 \sqrt{5} \sqrt{11}}{10 \times 4 \times 4}\\\\ &= 20 \sqrt{5} \sqrt{11}\\\\ &= 20 \sqrt{55} \end{aligned}\]
Evaluating numerically, \[\begin{aligned} R &= 20 \sqrt{55} \\&\approx 148.3 \text{ m}. \end{aligned}\] Thus, the maximum horizontal distance the ball can travel without touching the ceiling is approximately \(148.3 \text{ m}\).
Q13. A cricketer can throw a ball to a maximum horizontal distance of 100 m. How much high above the ground can the cricketer throw the same ball
Solution
The maximum horizontal distance (range) to which the cricketer can throw the ball is given as \( R_{\max} =
100 \text{ m} \).
For a projectile, the range is maximum when the angle of projection \( \theta = 45^{\circ}
\).
In that case, the maximum range is related to the initial speed \( u \) by
\[
R_{\max} = \frac{u^{2}}{g},
\]
where \( g \) is the acceleration due to gravity.
Substituting the given maximum range, \[\begin{aligned} 100 &= \frac{u^{2}}{g}\\\\ \Rightarrow u^{2} &= 100g \end{aligned}\] Taking \( g = 10 \text{ m s}^{-2} \), \[\begin{aligned} u^{2} &= 100 \times 10 \\&= 1000\\ \Rightarrow u &= 10\sqrt{10} \text{ m s}^{-1} \end{aligned}\]
The question now asks how high above the ground the cricketer can throw the same ball with this speed. For maximum vertical height, the ball must be thrown vertically upward. The maximum height \( H \) in vertical projection is \[ H = \frac{u^{2}}{2g} \] Using \( u^{2} = 1000 \) and \( g = 10 \text{ m s}^{-2} \), \[\begin{aligned} H &= \frac{1000}{2 \times 10}\\\\ &= \frac{1000}{20}\\\\ &= 50 \text{ m} \end{aligned}\]
Therefore, with the same initial speed, the cricketer can throw the ball up to a maximum height of \( 50 \text{ m} \) above the ground.
Q14. A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone ?
Solution
The stone is tied to a string of length \( r = 80 \text{ cm} = 0.8 \text{ m} \) and moves with uniform speed
in a horizontal circle.
It completes \(14\) revolutions in \(25 \text{ s}\).
Therefore, the time taken to complete one revolution is
\[
T = \frac{25}{14} \text{ s}
\]
The speed \( v \) of the stone in uniform circular motion is given by \[\begin{aligned} v &= \frac{2\pi r}{T}\\\\ &= \frac{2\pi r}{25/14}\\\\ &= 2\pi r \cdot \frac{14}{25} \end{aligned}\]
Substituting \( r = 0.8 \text{ m} \) and \( \pi \approx \frac{22}{7} \)
\[\begin{aligned} v &= 2 \times \frac{22}{7} \times 0.8 \times \frac{14}{25}\\\\ &= \frac{3.2 \times 22}{25}\\\\ &\approx 2.82 \text{ m s}^{-1} \end{aligned}\]The magnitude of centripetal acceleration in uniform circular motion is \[ a = \frac{v^{2}}{r} \]
Using \( v \approx 2.82 \text{ m s}^{-1} \) and \( r = 0.8 \text{ m} \)
\[\begin{aligned} a &= \frac{(2.82)^{2}}{0.8}\\\\ &= \frac{2.82 \times 2.82}{0.8}\\\\ &\approx 9.9 \text{ m s}^{-2} \end{aligned}\]The direction of this acceleration is along the radius of the circular path and always directed towards the centre of the circle (centripetal acceleration).
Q15. An aircraft executes a horizontal loop of radius 1.00 km with a steady speed of 900 km/h. Compare its centripetal acceleration with the acceleration due to gravity.
Solution
The aircraft is moving in a horizontal circle of radius \( r = 1.00 \text{ km} = 1000 \text{ m} \) with a
constant speed of \( 900 \text{ km h}^{-1} \).
To compute the centripetal acceleration, the speed is first
converted into SI units (m s\(^{-1}\)).
The speed in m s\(^{-1}\) is \[\begin{aligned} v &= 900 \times \frac{5}{18} \\ &= 250 \text{ m s}^{-1} \end{aligned}\] The centripetal acceleration for circular motion is \[ a = \frac{v^{2}}{r} \] Substituting the values, \[\begin{aligned} a &= \frac{250 \times 250}{1000}\\\\ &= \frac{62500}{1000}\\\\ &= 62.5 \text{ m s}^{-2} \end{aligned}\]
The acceleration due to gravity is \( g \approx 9.8 \text{ m s}^{-2} \). The ratio of the centripetal acceleration to \( g \) is \[\begin{aligned} \frac{a}{g} &= \frac{62.5}{9.8} \\\\ &\approx 6.4 \end{aligned}\] Therefore, the centripetal acceleration of the aircraft is about \( 6.4\ \mathrm{g} \), i.e. approximately \(6.4\) times the acceleration due to gravity.
Q16. Read each statement below carefully and
state, with reasons, if it is true or false :
(a) The net acceleration of a particle in
circular motion is always along the radius of
the circle towards the centre
(b) The velocity vector of a particle at a point
is always along the tangent to the path
of the particle at that point
(c) The acceleration vector of a particle in
uniform circular motion averaged over one
cycle is a null vector
Solution
(a) The net acceleration of a particle in circular motion is always along the radius of the circle towards
the
centre – False.
In non-uniform circular motion there are two components of acceleration: a radial (centripetal) component
towards the centre and a tangential component along the tangent when the speed changes, so the net
acceleration
is not purely radial in general.
(b) The velocity vector of a particle at a point is always along the tangent to the path of the particle at
that
point – True.
Velocity is defined as the rate of change of position, and in curvilinear motion the instantaneous direction
of
motion is along the tangent to the trajectory, so the velocity vector is tangential at each point.
(c) The acceleration vector of a particle in uniform circular motion averaged over one cycle is a null
vector –
True.
In uniform circular motion the acceleration is always directed towards the centre, but over one full
revolution
the vector changes direction continuously and returns to its initial value, so the vector sum over a cycle
is
zero and the average acceleration vector is the zero (null) vector.
Q17. The position of a particle is given by \(\vec{r}=3.0\ t\hat{i}+2.0\ t^2\hat{j} + 4.0\ \hat{k}\) m
where
\(t\) is in seconds
and the coefficients have the proper units for \(\vec{r}\) to be in metres.
(a) Find the \(\vec{v}\) and \(\vec{a}\) of the particle?
(b) What is the magnitude and direction of velocity of the particle at \(t = 2.0\) s ?
Solution
The position of the particle as a function of time is given by \[ \vec{r}(t) = 3.0t\,\hat{i} + 2.0t^{2}\,\hat{j} + 4.0\,\hat{k} \ \text{m}, \] where \( t \) is in seconds.
The velocity is the time derivative of the position vector: \[\begin{aligned} \vec{v}(t) &= \frac{d\vec{r}}{dt}\\\\ &= \frac{d}{dt}\left(3.0t\,\hat{i} + 2.0t^{2}\,\hat{j} + 4.0\,\hat{k}\right)\\\\ &= 3.0\,\hat{i} + 4.0t\,\hat{j} \end{aligned}\]
The acceleration is the time derivative of the velocity vector:
\[\begin{aligned} \vec{a}(t) &= \frac{d\vec{v}}{dt}\\\\ &= \frac{d}{dt}\left(3.0\,\hat{i} + 4.0t\,\hat{j}\right)\\\\ &= 4.0\,\hat{j}. \end{aligned}\]At \( t = 2.0 \text{ s} \), the velocity vector becomes \[\begin{aligned} \vec{v}_{(t=2)} &= 3.0\,\hat{i} + 4.0 \times 2.0\,\hat{j}\\ &= 3.0\,\hat{i} + 8.0\,\hat{j} \end{aligned}\] The magnitude of this velocity is \[\begin{aligned} \left|\vec{v}_{(t=2)}\right| &= \sqrt{3^{2} + 8^{2}}\\ &= \sqrt{9 + 64}\\ &= \sqrt{73}\\ &\approx 8.54 \ \text{m s}^{-1} \end{aligned}\]
The direction of the velocity at \( t = 2.0 \text{ s} \) is given by the angle \( \theta \) that \(\vec{v}(2)\) makes with the positive \(x\)-axis in the \(xy\)-plane: \[\begin{aligned} \tan\theta &= \frac{v_{y}}{v_{x}} \\\\ &= \frac{8}{3},\\\\ \theta &= \arctan\left(\frac{8}{3}\right) \\\\ &\approx 69.4^{\circ} \end{aligned}\] Thus, at \( t = 2.0 \text{ s} \) the particle has speed approximately \( 8.54 \ \text{m s}^{-1} \) directed \( \sim 69.4^{\circ} \) above the positive \(x\)-axis.
Q18. A particle starts from the origin at \(t = 0\) s with a velocity of \(10.0 \hat{j}\) m/s and moves
in
the x-y plane with a constant acceleration of \(8.0 \hat{i} + 2.0 \hat{j} \mathrm{m\ s^{-2}}\) .
(a) At what time is the x- coordinate of the particle 16 m? What is the y-coordinate of the particle at
that time?
(b) What is the speed of the particle at the time ?
Solution
The particle starts from the origin at \( t = 0 \) with initial velocity \[ \vec{v}_{0} = 10.0\,\hat{j}\ \text{m s}^{-1}, \] and moves under a constant acceleration \[ \vec{a} = 8.0\,\hat{i} + 2.0\,\hat{j}\ \text{m s}^{-2}. \]
The x–component of motion is governed by \[ x(t) = x_{0} + v_{0x} t + \tfrac{1}{2} a_{x} t^{2} \] Since the particle starts from the origin and the initial velocity has no x–component, \( x_{0} = 0 \) and \( v_{0x} = 0 \) Thus, \[\begin{aligned} x(t) &= \tfrac{1}{2} \cdot 8.0 \, t^{2} \\&= 4 t^{2} \end{aligned}\] For \( x = 16 \text{ m} \), \[\begin{aligned} 16 &= 4 t^{2}\\ \Rightarrow t^{2} &= 4\\ \Rightarrow t &= 2 \text{ s} \end{aligned}\] (taking the positive value since time is positive).
The y–coordinate at \( t = 2 \text{ s} \) is obtained from \[ y(t) = y_{0} + v_{0y} t + \tfrac{1}{2} a_{y} t^{2}, \] where \( y_{0} = 0 \), \( v_{0y} = 10.0 \text{ m s}^{-1} \), and \( a_{y} = 2.0 \text{ m s}^{-2} \). Therefore, \[\begin{aligned} y(2) &= 0 + 10.0 \times 2 + \tfrac{1}{2} \cdot 2.0 \times (2)^{2}\\\\ &= 20 + 1 \times 4\\\\ &= 24 \text{ m}. \end{aligned}\] Hence, when the x–coordinate is \(16 \text{ m}\), the y–coordinate is \(24 \text{ m}\).
The velocity at any time \( t \) is given by \[\begin{aligned} \vec{v}(t) &= \vec{v}_{0} + \vec{a} t\\ &= 10.0\,\hat{j} + (8.0\,\hat{i} + 2.0\,\hat{j}) t \end{aligned}\]
At \( t = 2 \text{ s} \),
\[\begin{aligned} \vec{v}(2) &= 10.0\,\hat{j} + (8.0\,\hat{i} + 2.0\,\hat{j}) \times 2\\ &= 10.0\,\hat{j} + 16.0\,\hat{i} + 4.0\,\hat{j}\\ &= 16.0\,\hat{i} + 14.0\,\hat{j}\ \text{m s}^{-1} \end{aligned}\]The speed of the particle at this instant is the magnitude of \(\vec{v}_{(t=2)}\): \[\begin{aligned} |\vec{v}(2)| &= \sqrt{16^{2} + 14^{2}}\\ &= \sqrt{256 + 196}\\ &= \sqrt{452}\\ &\approx 21.3 \ \text{m s}^{-1} \end{aligned}\] Thus, when the x–coordinate is \(16 \text{ m}\), the y–coordinate is \(24 \text{ m}\) and the speed of the particle is approximately \(21.3 \text{ m s}^{-1}\)
Q19. \(\hat{i}\) and \(\hat{j}\) are unit vectors along x- and y- axis respectively. What is the magnitude and direction of the vectors \(\hat{i}+\hat{j}\) , and \(\hat{i}-\hat{j}\) ? What are the components of a vector A= \(2 \hat{i} +3\hat{j}\) along the directions of \(\hat{i}+\hat{j}\) , and \(\hat{i}-\hat{j}\) ? [You may use graphical method]
Solution
The unit vectors \(\hat{i}\) and \(\hat{j}\) are directed along the positive x– and y–axes respectively. Consider first the vector \[ \vec{r} = \hat{i} + \hat{j}. \] Its components are \( (1, 1) \). The magnitude is \[\begin{aligned} |\vec{r}| &= \sqrt{1^{2} + 1^{2}} \\&= \sqrt{2}. \end{aligned}\] The direction angle \(\theta\) with respect to the positive x–axis is obtained from \[\begin{aligned} \tan\theta &= \frac{1}{1} \\&= 1 \\ \Rightarrow \theta &= 45^{\circ} \end{aligned}\] so \(\hat{i} + \hat{j}\) makes an angle of \(45^{\circ}\) in the first quadrant.
Now consider the vector \[ \vec{r}' = \hat{i} - \hat{j}, \] which has components \( (1, -1) \). Its magnitude is \[\begin{aligned} |\vec{r}'| &= \sqrt{1^{2} + (-1)^{2}} \\&= \sqrt{2}. \end{aligned}\] The direction angle is given by \[\begin{aligned} \tan\theta &= \frac{-1}{1} \\&= -1 \\ \Rightarrow \theta &= -45^{\circ} \end{aligned}\] so \(\hat{i} - \hat{j}\) makes an angle of \(45^{\circ}\) below the positive x–axis (in the fourth quadrant)
Let \[ \vec{A} = 2\hat{i} + 3\hat{j} \] To find the component of \(\vec{A}\) along the direction of \(\hat{i} + \hat{j}\), first note that the unit vector along \(\hat{i} + \hat{j}\) is \[ \hat{u}_{1} = \frac{\hat{i} + \hat{j}}{\sqrt{2}}. \] The component of \(\vec{A}\) along this direction is \[\begin{aligned} A_{1} &= \vec{A} \cdot \hat{u}_{1}\\\\ &= (2\hat{i} + 3\hat{j}) \cdot \frac{\hat{i} + \hat{j}}{\sqrt{2}}\\\\ &= \frac{2(\hat{i}\cdot\hat{i}) + 3(\hat{j}\cdot\hat{j})}{\sqrt{2}}\\\\ &= \frac{2 + 3}{\sqrt{2}}\\\\ &= \frac{5}{\sqrt{2}} \end{aligned}\]
Similarly, the unit vector along \(\hat{i} - \hat{j}\) is \[ \hat{u}_{2} = \frac{\hat{i} - \hat{j}}{\sqrt{2}} \] The component of \(\vec{A}\) along this direction is \[\begin{aligned} A_{2} &= \vec{A} \cdot \hat{u}_{2}\\\\ &= (2\hat{i} + 3\hat{j}) \cdot \frac{\hat{i} - \hat{j}}{\sqrt{2}}\\\\ &= \frac{2(\hat{i}\cdot\hat{i}) - 3(\hat{j}\cdot\hat{j})}{\sqrt{2}}\\\\ &= \frac{2 - 3}{\sqrt{2}}\\\\ &= -\frac{1}{\sqrt{2}} \end{aligned}\] Thus, \(\vec{A}\) has a component of magnitude \(\dfrac{5}{\sqrt{2}}\) along \(\hat{i} + \hat{j}\) and a component of magnitude \(\dfrac{1}{\sqrt{2}}\) along \(\hat{i} - \hat{j}\), the latter being directed opposite to \(\hat{i} - \hat{j}\)
Q20. For any arbitrary motion in space, which of the following relations are true :
(a) \(\vec{v}_\text{average}\) = (1/2) \((\vec{v}\)(t1 ) + \(\vec{v}\) (t2 ))
(b) \(\vec{v}_\text{average}\) = [\(\vec{r}\)(t2 ) - \(\vec{r}\)(t1 ) ] /(t2 – t1 )
(c) \(\vec{v}\) (t) = \(\vec{v}\) (0) + \(\vec{a}\) t
(d) \(\vec{r}\) (t) = \(\vec{r}\) (0) + \(\vec{v}\) (0) t + (1/2) \(\vec{a}\) t2
(e) \(\vec{a}_\text{average}\) =[ \(\vec{v}\) (t2 ) - \(\vec{v}\) (t1 )] /( t2 – t1 )
Solution
For an arbitrary motion in space, the following relations are considered:
(a) \( \vec{v}_{\text{avg}} = \dfrac{1}{2}\left[\vec{v}(t_{1}) + \vec{v}(t_{2})\right] \) – This is
false
in general. It is valid only when the acceleration is constant so that velocity
varies linearly with time.
(b) \( \vec{v}_{\text{avg}} = \dfrac{\vec{r}(t_{2}) - \vec{r}(t_{1})}{t_{2} - t_{1}} \) – This is
true.
It is the definition of average velocity over the time interval \([t_{1},
t_{2}]\),
independent of how the acceleration varies.
(c) \( \vec{v}(t) = \vec{v}(0) + \vec{a}\,t \) – This is not always true.
It holds
only when
the acceleration \(\vec{a}\) is constant in time.
(d) \( \vec{r}(t) = \vec{r}(0) + \vec{v}(0)t + \dfrac{1}{2}\vec{a}t^{2} \) – This is not always
true.
This kinematic equation is derived assuming constant acceleration; it fails for
arbitrary
time‑dependent acceleration.
(e) \( \vec{a}_{\text{avg}} = \dfrac{\vec{v}(t_{2}) - \vec{v}(t_{1})}{t_{2} - t_{1}} \) – This is
true.
It is the definition of average acceleration over the interval \([t_{1}, t_{2}]\)
for
any kind of motion.
Q21. Read each statement below carefully and state, with reasons and examples, if it is
true or false :
A scalar quantity is one that
(a) is conserved in a process
(b) can never take negative values
(c) must be dimensionless
(d) does not vary from one point to another in space
(e) has the same value for observers with different orientations of axes.
Solution
(a) A scalar quantity is one that is conserved in a process – This statement is
false.
Conservation is a separate physical property, not the definition of a scalar.
For
example, mass–energy (a scalar) is conserved in an isolated system, but temperature (also a scalar) is not
conserved; it can increase or decrease during a process.
(b) A scalar quantity can never take negative values – This statement is false.
Scalars can be positive, negative, or zero. For instance, electric potential at a point in space can be
negative, and scalar velocity (speed with sign chosen as per a convention in one‑dimensional motion) can
also take negative values.
(c) A scalar quantity must be dimensionless – This statement is false.
Many
scalars
have physical dimensions. Mass (kg), time (s), temperature (K), and energy (J) are all dimensional scalar
quantities. Dimensionless scalars (like refractive index) are only a subset of all scalar quantities.
(d) A scalar quantity does not vary from one point to another in space – This statement is
false.
A scalar field can vary from point to point. For example, temperature in a room
is a
scalar that may be higher near the ceiling and lower near the floor; similarly, electric potential varies
with position around a charge distribution.
(e) A scalar quantity has the same value for observers with different orientations of axes – This
statement is true.
The defining feature of a scalar is that it is independent of the
orientation of the coordinate system. Rotating the axes changes only the components of vectors, not the
value of scalars like mass, time interval, or temperature.
Q22.An aircraft is flying at a height of 3400 m above the ground. If the angle subtended at a ground observation point by the aircraft positions 10.0 s a part is 30°, wat is the speed of the aircraft ?
Solution
An aircraft flies horizontally at a constant height of \( 3400 \text{ m} \) above the ground. The angle between the two lines of sight from a ground observer to the aircraft at positions separated by \( 10.0 \text{ s} \) is \( 30^{\circ} \). The aircraft’s horizontal displacement in this time interval is the chord of the right triangle whose opposite side is the height and whose angle at the observer is \( 30^{\circ} \).
Let the horizontal distance travelled by the aircraft in \( 10.0 \text{ s} \) be \( x \). From the right‑angled triangle, \[\begin{aligned} \tan 30^{\circ} &= \frac{x}{3400}\\\\ \Rightarrow x &= 3400\ \tan 30^{\circ}\\\\ &= 3400 \times \frac{1}{\sqrt{3}} \end{aligned}\]
The speed \( v \) of the aircraft is the horizontal distance travelled divided by the time interval: \[\begin{aligned} v &= \frac{x}{t}\\\\ &= \frac{3400 / \sqrt{3}}{10}\\\\ &= \frac{3400}{10\sqrt{3}}\\\\ &\approx 196 \text{ m s}^{-1} \end{aligned}\] Thus, the speed of the aircraft is approximately \( 196 \text{ m s}^{-1} \).
