MOTION IN A STRAIGHT LINE-Execrise
Physics - Exercise
Q1. In which of the following examples of motion, can the body be considered
approximately a point object:
(a) a railway carriage moving without jerks between two stations.
(b) a monkey sitting on top of a man cycling smoothly on a circular track.
(c) a spinning cricket ball that turns sharply on hitting the ground.
(d) a tumbling beaker that has slipped off the edge of a table.
Solution
(a) and (b) allow the body to be considered approximately a point object.Reason
Railway Carriage (a)The railway carriage moves smoothly over large distances between stations, where its size is negligible compared to the path length. Thus, it approximates a point object.
Monkey on Cyclist (b)
The monkey's size is much smaller than the circular track's dimensions, making it a point object relative to the track. The smooth cycling supports this approximation.
Cricket Ball (c)
The spinning cricket ball's size compares to the sharp turning distance on the ground, preventing point object treatment.
Tumbling Beaker (d)
The beaker's size matches the table height during tumbling, so its extended motion invalidates point object approximation.
Q2. The position-time (x-t) graphs for two children A and B returning from their school
O to their homes P and Q respectively are shown in Fig. 2.9. Choose the correct
entries in the brackets below ;
(a) (A/B) lives closer to the school than (B/A)
(b) (A/B) starts from the school earlier than (B/A)
(c) (A/B) walks faster than (B/A)
(d) A and B reach home at the (same/different) time
(e) (A/B) overtakes (B/A) on the road (once/twice)
Solution
(a) A lives closer to the school than B. (b) A starts from the school earlier than B. (c) B walks faster than A. (d) A and B reach home at the same time. (e) B overtakes A on the road once.Q3. A woman starts from her home at 9.00 am, walks with a speed of \(5\ km\ h^{–1}\) on a straight road up to her office \(2.5\ km\) away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of \(25\ km\ h^{–1}\). Choose suitable scales and plot the x-t graph of her motion.
Solution
Q4. A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
Solution
Given:
Each step = 1 mTime per step = 1 s
Pattern: 5 steps forward, then 3 steps backward (repeated)
1. Net displacement per cycle
In one complete cycle:
Forward displacement \(= 5\mathrm{\ m}\)
Backward displacement \(= \mathrm{\ m}\)
Net displacement \(=5-3=2\mathrm{\ m}\)
Time for one cycle \(=5+3=8\mathrm{\ s}\)
So, after n complete cycles, displacement \(=2n\mathrm{\ m}\).
2. Cycles before the last forward run
The pit is 13 m from the start. He may fall during the forward 5 steps of some cycle, so the lastbackward 3 steps of that cycle will not occur.
During the last (incomplete) cycle, he can move at most 5 m forward before falling. Therefore the
distance covered only by complete cycles is:
\[13-5=8\mathrm{\ m}\]
Set this equal to displacement from n full cycles:
\[2n=8\Rightarrow n=4\]
So he completes 4 full cycles before the final forward run.
Time for 4 cycles \[=4\times8=32\mathrm{\ s}\]
At this instant he is 8 m from the start.
3. Final forward steps to the pit
From 8 m, he walks forward:After 1 step: 9 m (33 s)
After 2 steps: 10 m (34 s)
After 3 steps: 11 m (35 s)
After 4 steps: 12 m (36 s)
After 5 steps: 13 m (37 s) → falls into the pit.
Time for these 5 steps \(=5\mathrm{\ s}\)
4. Total time
\[\mathrm{Total\ time}=32\mathrm{\ s}+\mathrm{5\ s}=37\mathrm{\ s}\]So, the drunkard takes 37 s to fall into the pit 13 m away from the start
Q5. A car moving along a straight highway with speed of 126 km h–1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop ?
Solution
Speed of the car is given as \(126 \,\text{km h}^{-1}\). Converting this speed into SI units: \[\begin{aligned} 126 \,\text{km h}^{-1} &= \frac{126 \times 1000}{3600} \,\text{m s}^{-1}\\\\ &= \frac{126 \times 5}{18} \,\text{m s}^{-1}\\\\ &= 35 \,\text{m s}^{-1} \end{aligned}\]
The car is brought to rest in a distance of \(200 \,\text{m}\). Thus, initial velocity \(u = 35 \,\text{m s}^{-1}\), final velocity \(v = 0\), and displacement \(s = 200 \,\text{m}\). Using the kinematic equation \[ v^{2} - u^{2} = 2as, \] we substitute the values: \[ 0^{2} - (35)^{2} = 2 \times a \times 200. \] This gives \[\begin{aligned} -35^{2} &= 400a \\\\ \Rightarrow a &= \frac{-35 \times 35}{400} \\\\&= -3.06 \,\text{m s}^{-2} \,(\text{approximately}). \end{aligned}\]
The negative sign of \(a\) shows that the acceleration is opposite to the direction of motion, so the car has a uniform retardation of \(3.06 \,\text{m s}^{-2}\). To find the time taken to stop, use \[ v = u + at. \] Substituting \(v = 0\), \(u = 35 \,\text{m s}^{-1}\), and \(a = -3.06 \,\text{m s}^{-2}\), \[\begin{aligned} 0 &= 35 - 3.06\,t \\\\ \Rightarrow 3.06\,t &= 35 \\\\ \Rightarrow t &= \frac{35}{3.06} \\\\ &\approx 11.44 \,\text{s}. \end{aligned}\]
Therefore, the retardation of the car is \(3.06 \,\text{m s}^{-2}\) and the car takes approximately \(11.44 \,\text{s}\) to come to rest.
Q6. A player throws a ball upwards with an initial speed of 29.4 m s–1.
(a) What is the direction of acceleration during the upward motion of the ball ?
(b) What are the velocity and acceleration of the ball at the highest point of its motion ?
(c) Choose the x = 0 m and t = 0 s to be the location and time of the ball at its
highest point, vertically downward direction to be the positive direction of
x-axis, and give the signs of position, velocity and acceleration of the ball
during its upward, and downward motion.
(d) To what height does the ball rise and after how long does the ball return to the
player’s hands ? (Take g = 9.8 m s–2 and neglect air resistance).
Solution
(a) During the upward motion of the ball, the only acceleration acting on it is due to gravity. This acceleration always acts vertically downward, that is, towards the centre of the Earth, irrespective of whether the ball is moving up or down.
(b) At the highest point of its motion, the ball comes momentarily to rest before starting to fall back, so its instantaneous velocity is \(0 \,\text{m s}^{-1}\). However, the acceleration is still equal to the acceleration due to gravity and acts downward, so its magnitude remains \(9.8 \,\text{m s}^{-2}\) directed vertically downward.
(c) The origin \(x = 0 \,\text{m}\) and \(t = 0 \,\text{s}\) is chosen at the highest point, and the vertically downward direction is taken as positive. During the upward motion (i.e. when the ball is below the highest point and moving upward), the position of the ball is below the origin so \(x\) is negative, the velocity is directed upward so it is negative, while the acceleration is downward so it is positive. During the downward motion, the ball is still below the origin and moving downward, hence both position \(x\) and velocity are positive, and acceleration remains positive, since it is always directed downward.
(d) Initial velocity of projection is \(u = 29.4 \,\text{m s}^{-1}\) and acceleration due to gravity is taken as \(g = 9.8 \,\text{m s}^{-2}\) acting downward. For upward motion, taking upward as positive, the acceleration becomes \(a = -g = -9.8 \,\text{m s}^{-2}\). Let the maximum height reached be \(h\). At the highest point, the final velocity is \(v = 0\). Using the kinematic relation \[ v^{2} - u^{2} = 2 a h \] we substitute \(v = 0\), \(u = 29.4 \,\text{m s}^{-1}\) and \(a = -9.8 \,\text{m s}^{-2}\): \[ 0 - (29.4)^{2} = 2 \times (-9.8) \times h \] This gives \[\begin{aligned} h &= \frac{(29.4)^{2}}{2 \times 9.8}\\\\ &= \frac{29.4 \times 29.4}{2 \times 9.8}\\\\ &= 44.1 \,\text{m} \end{aligned}\] Thus, the ball rises to a height of \(44.1 \,\text{m}\) above the point of projection.
Time taken to reach the highest point is obtained from \[ v = u - g t \] At the top, \(v = 0\), so \[\begin{aligned} 0 &= 29.4 - 9.8 t\\\\ \Rightarrow 9.8 t &= 29.4\\\\ \Rightarrow t &= \frac{29.4}{9.8}\\\\ &= 3 \,\text{s} \end{aligned}\] The time of ascent equals the time of descent in this symmetric vertical motion with constant acceleration, so the total time of flight \(T\) is \[ T = 2t = 2 \times 3 = 6 \,\text{s} \] Therefore, the ball returns to the player’s hands after \(6 \,\text{s}\)
Q7. Read each statement below carefully and state with reasons and examples, if it is
true or false ;
A particle in one-dimensional motion
(a) with zero speed at an instant may have non-zero acceleration at that instant
(b) with zero speed may have non-zero velocity,
(c) with constant speed must have zero acceleration,
(d) with positive value of acceleration must be speeding up.
Solution
(a) True. A particle can have zero speed at an instant and still have non‑zero acceleration at that instant. For example, a ball thrown vertically upward has zero speed at the highest point, but its acceleration due to gravity is still \(\mathrm{9.8\ m\ s^{−2}}\) downward there.
(b) False. Speed is defined as the magnitude of velocity, so if the speed is zero, the magnitude of velocity is also zero and hence the velocity itself must be zero. Therefore, a particle with zero speed cannot have a non‑zero velocity.
(c) True (for one‑dimensional motion). If the motion is along a straight line and the speed is constant, the velocity (which in 1D is just speed with a sign) is also constant, so its rate of change is zero and the acceleration must be zero. A car moving on a straight highway at a steady \(\mathrm{60 km h^{−1}}\) is a standard example of motion with constant speed and zero acceleration.
(d) False. A positive acceleration means the velocity is increasing in the positive direction, but if the instantaneous velocity is negative, the object can actually be slowing down while the acceleration is positive. For instance, when a ball is thrown straight up and upward is taken as the positive direction, its velocity is positive while going up but its acceleration is negative; if instead downward is taken as positive, the acceleration becomes positive while the ball is still moving upward (negative velocity in that sign convention), so its speed is decreasing even though acceleration is positive.
Q8. A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.
Solution
Height from which the ball is dropped is \(90 \,\text{m}\)
Acceleration due to gravity is \(g = 9.8 \,\text{m s}^{-2}\)
Since the ball is dropped, its initial velocity is \(u = 0\)
Speed of the ball just before it hits the ground is obtained from \[ v^{2} - u^{2} = 2 g s \] Substituting \(u = 0\), \(g = 9.8 \,\text{m s}^{-2}\) and \(s = 90 \,\text{m}\), \[\begin{aligned} v^{2} &= 2 \times 9.8 \times 90 \\\\&= 19.6 \times 90, \end{aligned}\] so \[\begin{aligned} v &= \sqrt{196 \times 9} \\\\&= 14 \times 3 \\\\&= 42 \,\text{m s}^{-1} \end{aligned}\]
Time taken to reach the ground for the first time follows from \[ v = u + g t \]
Taking downward as positive, we put \(u = 0\), \(v = 42 \,\text{m s}^{-1}\), \(g = 9.8\ m\ s^{-2}\),
\[\begin{aligned} 42 &= 0 + 9.8 t \\\\\Rightarrow t &= \frac{42}{9.8} \\\\&\approx 4.3 \,\text{s} \end{aligned}\]At the first collision with the floor, the ball loses one tenth of its speed. Therefore, the speed just after the collision is \[\begin{aligned} v_{1} &= 42 - \frac{1}{10} \times 42 \\\\&= 42 - 4.2 \\\\&= 37.8 \,\text{m s}^{-1} \end{aligned}\] This speed is directed upward immediately after the bounce.
Time taken to reach the maximum height after the bounce (where its speed becomes zero) can be found from \[ v = u - g t \] Taking upward as positive for this part, \(u = 37.8 \,\text{m s}^{-1}\), \(v = 0\), \(g = 9.8 \,\text{m s}^{-2}\), so \[\begin{aligned} 0 &= 37.8 - 9.8 t \\&\Rightarrow 9.8 t \\&= 37.8 \\\Rightarrow t &= \frac{37.8}{9.8} \\&\approx 3.86 \,\text{s}. \end{aligned}\]
The time for the complete motion of this second segment (from just after the first bounce, up to maximum height, and then back to the ground with the same speed \(37.8 \,\text{m s}^{-1}\)) is \[\begin{aligned} T &= 2 t \\&= 2 \times 3.86 \\&= 7.72 \,\text{s} \end{aligned}\] Hence, the time from \(t = 0\) until the ball hits the ground for the second time is approximately \[ 4.3 + 7.72 = 12.02 \,\text{s} \] which matches the required interval from \(t = 0\) to about \(12 \,\text{s}\)
For the speed–time graph, the main points (using speed as a positive quantity) are:
at \(t = 0\), speed \(= 0 \,\text{m s}^{-1}\);
at \(t \approx 4.3 \,\text{s}\), speed \(= 42 \,\text{m s}^{-1}\);
immediately after the bounce, speed instantly changes to \(37.8 \,\text{m s}^{-1}\);
at \(t \approx 4.3 + 3.86 = 8.16 \,\text{s}\), speed \(= 0 \,\text{m s}^{-1}\);
and at \(t \approx 12.02 \,\text{s}\), just before the second impact, speed again becomes \(37.8
\,\text{m s}^{-1}\)
Q9. Explain clearly, with examples, the distinction between :
(a) magnitude of displacement (sometimes called distance) over an interval of time,
and the total length of path covered by a particle over the same interval;
(b) magnitude of average velocity over an interval of time, and the average speed
over the same interval. [Average speed of a particle over an interval of time is
defined as the total path length divided by the time interval]. Show in both (a)
and (b) that the second quantity is either greater than or equal to
the first. When is the equality sign true ? [For simplicity, consider one-dimensional motion only].
Solution
Magnitude of displacement and magnitude of average velocity are both based on the straight line change in position, while total path length and average speed use the entire route actually travelled, so the latter are always greater than or equal to the former in one dimension.
(a) Displacement vs total path length
Magnitude of displacement over a time interval is the straight line distance between the initial and final positions of the particle, ignoring the actual route in between. In one dimensional motion this is is simply \[\mid x_{\mathrm{final}}-x_{\mathrm{initial}}\mid\] The total path length is the full distance actually covered along the line, including any back and forth motion. For example, if a particle moves from x=0 m to x=+10 m and then back to x=+4 m, the magnitude of displacement isent is \mid4-0\mid=4 m, while the path length is 10+6=16 m. Thus, in 1D, \[\mathrm{path\ length}\geq\mathrm{magnitude\ of\ displacement}\]
with equality only when the motion is in a single direction with no reversals (for instance, from 0 m to +10
m
without coming back at all, where both are 10 m).
(b) Average velocity vs average speed
Magnitude of average velocity over a time interval \(\Delta t \) is
\[\mid \overline{v}∣=∣ΔxΔt∣\]where \(\Delta x\) is the displacement in that time. Average speed is defined as
\[\mathrm{average\ speed}=\frac{\mathrm{total\ path\ length}}{\Delta t}\]Using the same example as above (from 0 m to 10 m and back to 4 m in 7 s), the displacement is 4 m, so the magnitude of average velocity is \(\mid \overline{v}∣=4/7\ m\ s^{-1}\), while the path length is 16 m, giving an average speed of \(16/7\ m\ s^{-1}\), which is larger. In general, since path length \(\geq\) magnitude of displacement, dividing both by the same \(\Delta t\) gives \[\mathrm{average\ speed}\geq\mid \overline{v}∣\]
Equality holds only when path length equals the magnitude of displacement, i.e. when the particle moves along the line without ever reversing direction (for example, uniform motion from 0 m to 10 m; then both average speed and \(\mid \overline{v}∣\) are 10 m divided by the same time interval).
Q10. A man walks on a straight road from his home to a market 2.5 km away with
a speed of 5 km h–1. Finding the market closed, he instantly turns and
walks back home with a speed of 7.5 km h–1. What is the
(a) magnitude of average velocity, and
(b) average speed of the man over the interval of time (i) 0 to 30 min, (ii)
0 to 50 min, (iii) 0 to 40 min ?
Solution
Given:
Distance from home to market = 2.5 km
Speed from home to market = 5 km/h
Speed from market back to home = 7.5 km/h
Let’s break this down step by step.
(a) Magnitude of Average Velocity
The average velocity is defined as the total displacement divided by the total time taken.Displacement:
The displacement is a straight line from the initial point (home) to the final point (home again), which is zero since he returns to his starting point. So, the magnitude of average velocity is: \[\begin{aligned} \mathrm{Average\ velocity}&=\frac{\mathrm{Total\ displacement}}{\mathrm{Total\ time}}\\\\&=\frac{0}{\mathrm{Total\ time}}\\\\&=0\end{aligned}\] Thus, the magnitude of average velocity is 0.
(b) Average Speed
Average speed is defined as the total distance traveled divided by the total time taken.Total Distance:
Distance from home to market = 2.5 km
Distance from market back to home = 2.5 km
So, total distance = 2.5+2.5=5km
Time Taken:
Time to go from home to market:
\[\begin{aligned} \mathrm{Time}&=\frac{\mathrm{Distance}}{\mathrm{Speed}}\\\\&=\frac{2.5}{5}\\\\&=0.5\mathrm{\ hours}\end{aligned}\] Time to go from market back to home:
\[\begin{aligned} \mathrm{Time}&=\frac{\mathrm{Distance}}{\mathrm{Speed}}\\\\&=\frac{2.5}{7.5}\\\\&=\frac{1}{3}\mathrm{\ hours} \end{aligned}\] \[\begin{aligned}\text{Total time }&= 0.5+\frac{1}{3}\\\\&=\frac{3}{6}+\frac{2}{6}\\\\&=\frac{5}{6}\ \mathrm{hours}\end{aligned}\]
Now let’s calculate the average speed over the given intervals.
(i) 0 to 30 minutes (0 to 0.5 hours)
In this case, the man has walked only part of the way, and by 30 minutes, he is still on his way to the market.Distance traveled by 30 minutes: \[\begin{aligned} \mathrm{Speed}\times\mathrm{Time}&=5\times0.5\\&=2.5\ \mathrm{km} \text{ (reached the market)} \end{aligned}\] Time: 0.5 hours = 30 minutes
Since he has not yet walked back, the total distance is still 2.5 km.
So, the average speed in this interval: \[ \begin{aligned} \mathrm{Average\ speed}&=\frac{\mathrm{Total\ distance}}{\mathrm{Total\ time}}\\\\&=\frac{2.5}{0.5}\\\\&=5\mathrm{\ km/h} \end{aligned} \]
(ii) 0 to 50 minutes (0 to 5/6 hours)
In 50 minutes (5/6 hours), the man has walked both to the market and back home, but he hasn't yet completed
the entire journey.
Time to reach the market: 0.5 hours (covered 2.5 km).
In the remaining \(\frac{1}{3}\) hours, he has traveled \[7.5\times\frac{1}{3}=2.5\ \mathrm{km}\] which means he’s halfway back home.
Thus, by 50 minutes, he’s covered 2.5 km towards the market, and 2.5 km back towards home, totaling 5 km. The total time taken is 50 minutes or \(\frac{5}{6}\) hours.
The total distance is still 5 km, and the time is \(\frac{5}{6}\) hours.
So, the average speed over this interval is: \[ \begin{aligned} \mathrm{Average\ speed}&=\frac{\mathrm{Total\ distance}}{\mathrm{Total\ time}}\\\\&=\frac{5}{\frac{5}{6}}\\\\&=6\mathrm{\ km/h} \end{aligned} \](iii) 0 to 40 minutes (0 to 2/3 hours)
In 40 minutes, or \(\frac{2}{3}\) hours, the man has:
Covered \(5\times\frac{2}{3}=3.33\) km in total.
He walked to the market, which is 2.5 km in 0.5 hours.
In the remaining \[\begin{aligned}\frac{2}{3}-\frac{1}{2}&=\frac{1}{6}\ \mathrm{hours}\end{aligned}\] he covers \[7.5\times\frac{1}{6}=1.25\ \mathrm{km}\] on his way back home.
So, by 40 minutes, he has covered 2.5 km to the market and 1.25 km on his way back, for a total of 3.75 km.
Total time is \(\frac{2}{3}\) hours, or 40 minutes.Thus, the average speed over this interval is: \[\begin{aligned} \mathrm{Average\ speed}&=\frac{\mathrm{Total\ distance}}{\mathrm{Total\ time}}\\\\&=\frac{3.75}{\frac{2}{3}}\\\\&=5.625\mathrm{\ km/h} \end{aligned}\]
Summary:
(a) Magnitude of Average Velocity: \(0\ \mathrm{km/h}\)
(b) Average Speed:
(i) 0 to 30 min: \(5\ \mathrm{km/h}\)
(ii) 0 to 50 min: \(6\ \mathrm{km/h}\)
(iii) 0 to 40 min: \(5.625\ \mathrm{km/h}\)
Q11. In Exercises 2.9 and 2.10, we have carefully distinguished between average speed and magnitude of average velocity. No such distinction is necessary when we consider instantaneous speed and magnitude of velocity. The instantaneous speed is always equal to the magnitude of instantaneous velocity. Why?
Solution
Instantaneous speed equals the magnitude of instantaneous velocity because, in an infinitesimally small time interval, distance travelled and magnitude of displacement become the same.
In a finite time interval, distance can exceed displacement if the particle turns around or follows a curved path, which is why average speed is generally greater than or equal to the magnitude of average velocity. But instantaneous quantities are defined as limits over a very small time interval \({\Delta}{t}\rightarrow{0}\). In that tiny interval the particle does not have time to reverse direction appreciably, so its path segment is effectively straight, and the path length \({\Delta}s\) equals the magnitude of displacementent \(\mid{\Delta}{x}\mid\).
Mathematically, instantaneous speed is
\[{v}_{\mathrm{speed}}=\lim_{\Delta t \to 0}\dfrac{\Delta s}{\Delta t}\] and instantaneous velocity is the vector \[\vec{v}=\lim_{\Delta t to 0} \dfrac{\Delta \vec{x}}{\Delta t}\] whose magnitude is \[\mid\vec{v}\mid=\lim_{\Delta t to 0} \dfrac{\Delta \vec{x}}{\Delta t}\]limits are equal, so instantaneous speed \(=\mid\vec{v}\mid\) at every instant.
For example, a car going around a circular track at varying speed may have average speed \(40\ km\ h^{-1}\) and average velocity 0 over a full lap, but at any particular instant its speedometer reading (say \(36\ km\ h^{-1}\)) is exactly the magnitude of its instantaneous velocity vector along the tangent to the circle.
Q12. Look at the graphs (a) to (d) (Fig. 2.10) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.
Solution
None of the four graphs can represent one‑dimensional motion of a particle.
Graph (a): x–t graph
In one‑dimensional motion, the particle can have only one position at a given instant of time. In graph (a), the x–t curve crosses itself, meaning that at the same time \(t\) the particle would be at two different positions \(x\), which is impossible.
Graph (b): v–t graph
For a given instant of time, a particle in one dimension can have only a single value of velocity. Graph (b) is a circle in the v–t plane, so for one value of \(t\) there are two different velocities, which cannot occur in one‑dimensional motion.
Graph (c): speed–t graph
Speed is a scalar quantity and is defined as the magnitude of velocity, so it cannot be negative. In graph (c), the curve goes below the time axis, implying negative speed at some instants, which is unphysical.
Graph (d): total path length–t graph
Total path length is the cumulative distance travelled from the start; it can stay constant (if the particle is at rest) or increase with time, but it can never decrease. In graph (d), the line slants downward in the middle, showing total path length decreasing with time, which is impossible.
Q13. Figure 2.11 shows the x-t plot of one dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t>0 ? If not, suggest a suitable physical context for this graph.
Solution
It is not correct to say that the particle moves on a straight line for \(t < 0 \) and on a parabolic path for \(t>0\).
Why this interpretation is wrong
The graph is an x-t graph, not a path diagram.
The horizontal axis is time, not a spatial coordinate, so the curve cannot tell whether the spatial
path is straight or curved. It only shows how the position coordinate \(x\) changes with time along
some one dimensional line.
Saying “straight line path” or “parabolic path” refers to the shape of the trajectory in space (e.g. \(y \text{ vs }x\)), whereas Fig. 2.11 is only position vs time in one dimension. Suitable physical context for this graph
From the figure, for \(t< 0\) the plot is horizontal, so \(x\) is constant and the particle is at rest. For \(t>0,\ x\) increases with time with increasing slope, indicating motion with increasing velocity, i.e. motion with uniform acceleration starting from rest at \(t=0\).
A suitable example is:
A body initially at rest on a horizontal surface (for \(t< 0\)), and At \(t=0\), a constant unbalanced force begins to act on it so that it moves along a straight line with constant acceleration (for \(t> 0\)). Another common context is an object starting from rest and falling freely under gravity along a vertical line once it is released at \(t= 0\).
Q14. A police van moving on a highway with a speed of 30 km h–1 fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h–1. If the muzzle speed of the bullet is 150 m s–1, with what speed does the bullet hit the thief’s car ? (Note: Obtain that speed which is relevant for damaging the thief’s car).
Solution
Step 1: Convert vehicle speeds to \(m\ \mathrm{s^{-1}}\)
\[\begin{aligned} v_p &=30\ \mathrm{km\ h^{-1}}\\ &=\frac{30\times 1000}{3600}\\ &\approx 8.33\ \mathrm{m\ s^{-1}}\\\\ v_t &=192\mathrm{km\ h^{-1}}\\ &=\frac{192 \times 1000}{3600}\\& \approx 53.33\ \mathrm{m\ s^{-1}}\end{aligned}\]Here \(v_p\) is police van speed, \(v_t\) is thief’s car speed.
Step 2: Bullet speed relative to groundMuzzle speed \(=150\mathrm{m\ s^{-1}}\) is the speed of the bullet relative to the van.
So bullet speed relative to ground:
\[ \begin{aligned} v_b &=v_{\mathrm{muzzle}}+v_p\\&=150+8.33\\&=158.33\ \mathrm{m\ s^{-1}} \end{aligned} \]Step 3: Speed of bullet relative to thief’s car Both bullet and thief’s car move in the same direction, so relative speed is the difference:
\[\begin{aligned}v_{\mathrm{bullet\ w.r.t\ car}}&=v_b-v_t\\&=158.33-53.33\\&=105\ \mathrm{m\ s^{-1}} \end{aligned}\]This \(105\ m\ s^{-1}\) is the impact speed relevant for damaging the thief’s car.
Q15. Suggest a suitable physical situation for each of the following graphs (Fig 2.12):
Solution
One suitable physical situation can be given for each graph as follows.
Graph (a): x–t graph
The particle initially moves with constant velocity in the positive \(x\)-direction, then comes momentarily to rest, and finally moves with constant velocity in the negative \(x\)-direction. This happens, for example, when a ball lying on a smooth floor is kicked towards a wall, comes to rest for an instant on hitting the wall, and then rebounds back with smaller uniform speed.
Graph (b): v–t graph
The magnitude of velocity decreases in steps while its sign alternates, indicating repeated rebounds with successively smaller speeds. A suitable example is a ball dropped on a hard floor: it hits the floor with some downward velocity, rebounds with smaller upward velocity, falls again with reduced speed, and so on, until it eventually comes to rest.
Graph (c): a–t graph
The acceleration is zero for most of the time, but shows a sharp, short duration positive peak. This corresponds to a brief impulsive force acting on an otherwise uniformly moving or stationary body. A good example is a cricket ball moving horizontally and then being struck by a bat: during the short interval of contact, the ball experiences a large acceleration, while before and after impact the acceleration is negligible (approximately zero).
Q16. Figure 2.13 gives the x-t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter 13). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s.
Solution
At all these instants the motion is simple harmonic, so \(a=−ω^2\ x\): acceleration always has sign opposite to position.
At \(t=0.3\ s\): position \(x < 0\); the graph is sloping downwards (more negative with time), so velocity \(v < 0\); hence acceleration \(a>0\).
At \(t=1.2\ s\): position \(x>0\); the graph is rising (more positive with time), so velocity \(v>0\); therefore acceleration \(a < 0\)
At \(t=− 1.2\ s\): position \(x < 0\); going forward in time from \(− 1.2\ s\) towards 0 the graph moves upwards, so velocity is positive \(v>0\); with \(x < 0\), acceleration is again \(a> 0\).
Q17.Figure 2.14 gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least ? Give the sign of average velocity for each interval.
Solution
Average speed is greatest in interval 3 and least in interval 2.
Average speed in a given interval is proportional to the magnitude of the slope of the \(x-t\) graph in that interval. The slope is steepest (largest magnitude) in interval 3 and shallowest in interval 2, so interval 3 has the highest average speed and interval 2 the lowest.
The sign of average velocity is the sign of the slope of the \(x-t\) graph: it is positive in intervals 1 and 2 (graph rising with time) and negative in interval 3 (graph falling with time).
Q18. Figure 2.15 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest ? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D ?
Solution
Average acceleration has greatest magnitude in interval 2, and average speed is greatest in interval 3.
In a speed–time graph, the average acceleration over an interval equals the slope of the chord joining the endpoints of that interval. Since the curve falls most steeply between \(t = 2\ \text{s}\) and \(t = 3\ \text{s}\) (from B to C), the magnitude \(|\bar a|\) is largest in interval 2.
The average speed in an interval is effectively the mean height of the speed–time curve over that interval. Because the speeds are overall highest between \(t = 2\ \text{s}\) and \(t = 3\ \text{s}\), the average speed is greatest in interval 3.
Motion is along a fixed positive direction and the graph is of speed, so the velocity \(v\) is positive in all three intervals. Thus, for intervals 1, 2 and 3 we have \(v > 0\) throughout.
The sign of acceleration \(a\) follows from whether the speed is increasing or decreasing: in interval 1 (A to B) the speed increases, so \(a > 0\); in interval 2 (B to C) the speed decreases, so \(a < 0\); and in interval 3 (C to D) the speed increases again, so \(a > 0\).
At the labelled points, the accelerations are: at A, the curve is gently rising so the instantaneous acceleration is positive but small \((a > 0)\); at B, the speed has a local maximum so the slope of the tangent is zero and hence \(a = 0\); at C, the speed has a local minimum so again the slope is zero and \(a = 0\); at D, the point is at (or very near) the top of the next peak, so exactly at the highest point the slope is zero and \(a = 0\), with \(a > 0\) just before D and \(a < 0\) just after D.
