Q1. An object experiences a net zero external unbalanced force. Is it possible for the object to be travelling with a non-zero velocity? If yes, state the conditions that must be placed on
Solution:Yes, an object can have a non-zero velocity with zero net external force.
Let external force \(F\Rightarrow\) By Newton's second law of Motion, F can be written as $$\begin{aligned}F=ma\\ 0=ma\\ \Rightarrow a=0\end{aligned}$$ \(\Rightarrow\) acceleration zero does not mean velocity is zero, in the absence of an External force, an object may move with a constant velocity (Law of Inertia).
Q2. When a carpet is beaten with a stick, dust comes out of it. Explain.
Explanation:Due to inertia, when the carpet is beaten with a stick, it starts to move, but the dust particles on it tend to remain at rest. Because the dust does not immediately follow the motion of the carpet, it loses contact with the carpet and comes off its surface.
Q3. Why is it advised to tie any luggage kept on the roof of a bus with a rope?
Explanation:It is advised to tie any luggage kept on the roof of a bus with a rope because if the bus starts moving suddenly or the driver applies the brakes, the bus will change its motion, but the luggage will tend to resist this change due to inertia, which may cause it to fall off the roof.
Q4. A batsman hits a cricket ball, which then rolls on a level ground. After covering a short
distance, the
ball comes to rest. The ball slows to a stop because
(a) The batsman did not hit the ball hard enough.
(b) Velocity is proportional to the force exerted on the ball.
(c) There is a force on the ball opposing the motion.
(d) There is no unbalanced force on the ball, so the ball would want to come to rest.
(c) There is a force on the ball opposing the motion.\(\quad\Rightarrow\quad\color{blue} Correct\)
On the ground, friction acts opposite to motion. Also, air resistance resists motion. These opposing forces gradually reduce the velocity to zero.
Q5. A truck starts from rest and rolls down a hill with a constant acceleration. It travels a distance of 400 m in 20 s. Find its acceleration. Find the force acting on it if its mass is 7 tonnes (Hint: 1 tonne = 1000 kg.)
Solution:Using First Equation of Motion \((u=0,~t=20s)\) $$\begin{aligned}v&=u+at\\v&=0+20a \end{aligned} $$ $$v=20a\tag{1}$$ Uing third equation of Motion \(v^2 - u^2 = 2as\) $$\begin{aligned}v^{2}-u^{2}&=2as \\ v^{2}-0^{2}&=2\cdot a\cdot 400\\ \end{aligned}$$ $$v^{2}=800a\tag{2}$$ putting value of \(v\) from equation (1) to equation (2) $$\begin{aligned}\left( 20a\right) ^{2}&=800a\\ 400a^{2}&=800\\ a&=\dfrac{800}{400}\\ a&=2 m/s^2\end{aligned}$$ Force on th truck of mass \((m)=7 ~tonnes\) can be given as $$\begin{aligned} F&=ma\\ &=7000\times 2\\ &=14000 N\end{aligned}$$
Q6. A stone of 1 kg is thrown with a velocity of 20 m s–1 across the frozen surface of a lake, and comes to rest after travelling a distance of 50 m. What is the force of friction between the stone and the ice?
Solution:$$\begin{aligned}m&=1kg\\ u&=20ms^{-1}\\ s&=50m\\ v&=0\end{aligned}$$ Using the third equation of Motion \(v^2-u^2=2as\) $$\begin{aligned} v^{2}&=u^{2}+2as\\ 0^{2}&=\left( 20\right) ^{2}+2\cdot a\cdot 50\\ \Rightarrow 100a&=-20\times 20\\ a&=-4\times m/s^{2} \end{aligned}$$ $$\color{blue}\text{(-ve sign shows negative acceleration }\Rightarrow \text{retardation)}$$ Frictional force between the stone and ice can be given as $$\begin{aligned} F&=ma\\ 4&=1\cdot (-4)\\ &=-4N\end{aligned}$$
Q7. An 8000 kg engine pulls a train of 5 wagons, each of 2000 kg, along a horizontal track. If
the engine
exerts a force of 40000 N, and the track offers a friction force of 5000 N, then
calculate:
(a) the net accelerating force and
(b) The acceleration of the train.
\(\begin{aligned}\text{Wt of Engine} &= 8000kg\\ \text{Wt of one wagon} &= 2000kg\\ \text{therefore, wt of 5 Wagon} = 2000 \times 5 &= 10,000kg\\ \text{total wt of train = 10000 + 8000 }&=18000kg\\ \text{Force Exerted by Engine}&=40000N\\ \text{Friction Force}&=5000N\\ \text{Net Force in moving direction = 40000 - 5000} &=35000N\\ \text{Total Mass }M&=18000kg\\ \text{Net Accelerating Force}&=35000N \end{aligned}\\\) The acceleration of the train can be given as $$\begin{aligned} F&=ma\\ \Rightarrow a&=F/m\\ &=\dfrac{35000}{18000}\\ &=\dfrac{35}{18}\\ &\approx 1.944ms^{-2}\end{aligned}$$
Q8. An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of \(1.7m/s^2\) ?
Solution:Mass \((m)\) of automobile = 1500kg $$\begin{aligned} a &= 1.7m/s^2 \text{ -(given)}\end{aligned}$$ The force between the vehicle and the road may be given as $$\begin{aligned} F&= ma\\ &= 1500 \times 1.7\\ &= 2550.0\\ &= 2550N\end{aligned}$$
Q9. What is the momentum of an object of mass \(m\), moving with
a velocity \(v\)?
(a) \((mv)^2\\\)
(b) \(mv^2\\\)
(c) \(\frac{1}{2} mv^2\\\)
(d) \(mv\)
(d) \(p=mv\Rightarrow\) is Correct
Q10. Using a horizontal force of 200 N, we intend to move a wooden cabinet across a floor at a constant velocity. What is the friction force that will be exerted on the cabinet?
Solution:The friction force exerted on an object moving at constant velocity is equal in magnitude and opposite in direction to the applied force.
Since the cabinet is moving at constant velocity under a horizontal force of 200 N, the friction force must be 200 N, opposing the motion.
Q11. According to the third law of motion, when we push on an object, the object pushes back on us with an equal and opposite force. If the object is a massive truck parked along the roadside, it will probably not move. A student justifies this by answering that the two opposite and equal forces cancel each other. Comment on this logic and explain why the truck does not move.
Explanation:The student’s logic that “the two opposite and equal forces cancel each other” is incorrect because the two forces referred to in Newton’s third law act on different objects, not on the same object. According to Newton’s third law, if you push on the truck, the truck pushes back on you with an equal and opposite force. These forces are action-reaction pairs acting on two separate bodies— the push is on the truck, and the reaction is on the person pushing.
The reason the truck does not move is not because these two forces cancel out on the truck itself, but because the net force on the truck is zero. The truck is at rest because the forces acting on it (like friction between the truck’s wheels and the ground, or the truck’s own weight supported by the ground) balance the applied force, resulting in no acceleration.
Q12. A hockey ball of mass 200 g travelling at 10 \(m s^{-1}\) is struck by a hockey stick so as to return it along its original path with a velocity at 5 \(m s^{–1}\). Calculate the magnitude of the change of momentum occurred in the motion of the hockey ball by the force applied by the hockey stick.
Solution: Given that:Mass of the hockey ball \(m=200g\\\) Initial Velocity \(u=10m/s\\\) Final Velocity \(v=-5m/s\) as ball will move in opposite direction
Change in momentum \(\Delta p = m\Delta v\) $$\begin{aligned} \triangle p&=m\Delta v\\ &=\dfrac{200}{1000}\left( u-\left( -v\right) \right) & \color{blue}\left(\because\quad 200 g = \frac{200}{1000} kg\right) \\ &=\dfrac{200}{1000}\left( 10-\left( -5\right) \right) \\ &=\dfrac{200}{1000}\left( 10+5\right) \\ &=\dfrac{200\times 15}{1000}\\ &=\dfrac{200\times 15}{1000}\\ &=\dfrac{3000}{1000}\\ &=3N\end{aligned}$$
Q13. A bullet of mass 10 g travelling horizontally with a velocity of 150 m s–1 strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.
Solution:Mass of bullet \((m)=10 g\\\) Initial Velocity \(u = 150m/s\\\) Time when bullet comes to rest \(t = 0.03s\\\) Final Velocity of bullet \(v = 0\\\) Acceleration of the bullet can be calculated by using First Equation of Motion $$\begin{aligned} v&=u+at\\ 0&=150+a\cdot 0.03\\ -0.03a&=150\\\\ a&=-\dfrac{150}{0\cdot 03}&\color{blue}\text{(-ve value shows retardation)}\\\\ a&=-\dfrac{150\times 100}{3}\\\\ =&-5000m/s\end{aligned}$$ Force Exerted on the wooden block $$\begin{aligned} F&=ma\\ &=10\times 10^{-3}\times 5\times 10^{3}\\ &=50N\end{aligned}$$ Let Distance penetrated by the bullet be \(s\), therefore, using third equation of motion we ca find the distance $$\begin{aligned} v^{2}-u^{2}&=2as\\\\ 0-\left( 150\right) ^{2}&=2\times (-5\times 10^{3})\cdot s\\\\ \Rightarrow s&=\dfrac{150\times 150}{2\times 5\times 1000}\\\\ s&=2.25m\end{aligned}$$
Q14. An object of mass 1 kg travelling in a straight line with a velocity of \(10 m s^–1\) collides with, and sticks to, a stationary wooden block of mass 5 kg. Then they both move off together in the same straight line. Calculate the total momentum just before the impact and just after the impact. Also, calculate the velocity of the combined object.
Solution:Mass \((m)\) of the object = 1kg
Initial velocity \(u= 10m/s\)
Mass of wood \(M = 5kg\)
Total mass $$\begin{aligned}&= m + M\\ &= 1kg + 5kg\\ &= 6kg\end{aligned}$$ Moment just before the Impact $$\begin{aligned}p &= m u\\ &= 1\times 10\\ &= 10 kg m/s\end{aligned}$$ By conservation of momentum $$\begin{aligned} p_{1}&=p_{2}\\ m_1u&=m_2v&\left(m_1=m\Rightarrow1kg,~m_2=m+M\Rightarrow 6kg\right)\\ 1\cdot 10&=6.v\\\\ v_{2}&=\dfrac{10}{6}\\\\&=\dfrac{5}{3}\\\\ &\approx1\cdot 66m/s \end{aligned}$$ Moment just after the Impact $$\begin{aligned}p &= (m+M) v\\\\ &= 6\times \frac{5}{3}\\\\ &= 10 kg m/s\end{aligned}$$ By conservation of momentum
Q15. An object of mass 100 kg is accelerated uniformly from a velocity of \(5 m s^–1\) to \(8 m s^–1 \)in \(6 s\). Calculate the initial and final momentum of the object. Also, find the magnitude of the force exerted on the object
Solution:Mass of the object \((m) = 100 kg\\\) Initial velocity \(u = 5m/s\\\) Final velocity \(v = 8 m/s\\\) Time taken to accelerate \(t = 6s\\\) By first equation of motion $$\begin{aligned}v&=u+at\\ 8&=5+a\cdot 6\\ 8-5&=6a\\ \Rightarrow 6a&=3\\ a&=3/6\\ =&0.5ms^{-2}\end{aligned}$$ Initial momentum $$\begin{aligned}p &= mu\\ &= 100\times 5\\ &= 500 Kgm/s\end{aligned}$$ Final Momentum $$\begin{aligned}p &= mv\\ &= 100\times 8\\ &= 800 Kgm/s\end{aligned}$$ Force Exerted on the object: $$\begin{aligned} F&=ma\\&= 100\times 0.5\\&=50N\end{aligned}$$
Q16. Akhtar, Kiran and Rahul were riding in a motorcar that was moving with a high velocity on an expressway when an insect hit the windshield and got stuck on the windscreen. Akhtar and Kiran started pondering over the situation. Kiran suggested that the insect suffered a greater change in momentum as compared to the change in momentum of the motorcar (because the change in the velocity of the insect was much more than that of the motorcar). Akhtar said that since the motorcar was moving with a larger velocity, it exerted a larger force on the insect. And as a result the insect died. Rahul while putting an entirely new explanation said that both the motorcar and the insect experienced the same force and a change in their momentum. Comment on these suggestions.
Comments:Kiran’s suggestion that the insect suffered a greater change in momentum than the motorcar is incorrect because the change in momentum of the insect and the motorcar must be equal in magnitude due to the conservation of momentum, though the insect’s velocity changes much more since it has a much smaller mass than the motorcar.
Akhtar’s suggestion that the motorcar exerts a larger force on the insect because of its larger velocity is also wrong. By Newton’s third law, the forces on the insect and motorcar during the collision are equal in magnitude and opposite in direction. The insect dies not because of a larger force on it but because its small mass and sudden velocity change cause a large acceleration and damage.
Rahul’s explanation that both the motorcar and the insect experience the same force and the same magnitude of change in momentum is correct. However, the directions of these momentum changes are opposite. The motorcar’s large mass means a very small velocity change, while the insect’s small mass means a larger velocity change. mass leads to a large velocity change upon impact, causing fatal injury to the insect.
How much momentum will a dumbbell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be \(10 m s^–2\)
Solution:Given that:
Initial Velocity \(u=0\\\) Downward aceleration \(=10m/s^2\\\) Using third equation of Motion $$\begin{aligned}v^{2}-u^{2}&=2as\\ v^{2}-0^{2}&=2\cdot 10\cdot 0.8\\ v^{2}&=2\times 8\\ v&=\sqrt{16}\\ v&=4m/s\end{aligned}$$ therefore, Momentum of the dumbell $$\begin{aligned}p&=mv\\&=10\times 4\\&=40kgm/s\end{aligned}$$
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Frequently Asked Questions
Exercise
1. An athlete completes one round of a circular track of diameter 200 m in 40 s. What will be the distance covered and the displacement at the end of 2 minutes 20 s?
Solution:
Diameter (d) of the circular path = 200m
Hence, Circumference (s) of circular path = \(
s=\pi\times d\ \\\quad=\frac{22}{7}\times 20
\\\)
Time taken to complete one round = 40s
Hence, speed (\(v\)) of Athelete \(\\\\\Rightarrow
\require{cancel}
v=\dfrac{distance}{time}\\\\\quad= \dfrac{\frac{22}{7}\times 200}{40}
\\\\\quad=\dfrac{22\times200}{7\times40}\\\\\quad=\dfrac{22\times\cancelto{5}{200}}{7\times\cancelto{1}{40}}\\\\\quad=\dfrac{110}{7}\\\)
Distance coverd by athlete in time\(t\) 2 minutes and 20 seconds =\(
t=(2\times60) + 20=120+20=140~seconds
\)
Distance covered in time \\\(t\)=\(\require{cancel} speed\times
time\\\quad=v\times t \\\\\quad=\frac{110}{7}\times 140\\\\\quad=\frac{110}{\cancelto{1}7}\times
\cancelto{20}{140}\\\\\quad=2200 m\\\)
Total Number of rounds completed by athlete in 2 minutes and 20 seconds = \(
\require{cancel}
\\\dfrac{total~run ~time}{time ~taken ~to ~complete~ one
~roud}\\\\\quad=\dfrac{140}{40}\\\\\quad=\dfrac{\cancelto{7}{140}}{\cancelto{2}{40}}\\\\\quad=\frac{7}{2}\\\\\quad=3\dfrac{1}{2}\\\)It is to be noted that upon completion of every round, the Athlete will return to the same place where they started, which means displacement will be zero after completing one round.
Therefore, Total displacement will be the shortest distance (AB in Fig. 7.1) when he will
be completing \(\dfrac{1}{2}\) round, that is equal to diameter of the circular path=40m
fig 7.1
2. Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes and 30 seconds seconds and then turns around and jogs 100 m back to point C in another 1 minute. What are Joseph’s average speeds and velocities in jogging (a) from A to B and (b) from A to C?
Solution:
Time taken by Joseph to complete (A to B) 300m
\( \\\quad= \text{2 minute 30 second} \\\\\quad= (2×60) +30 \\\\\quad= 150 \text{ seconds} \) Average speed in jogging from A to B \(\\\quad= \text{total distance}/\text{total time}\\\quad\quad = 300/150\\\quad\quad = 2 m/s\\\) Speed During jogging, the distance A to Ctotal Distance \( \\\quad= AB + BC \\\\\quad= 300 + 100 \\\\\quad= 400 \\\) Total time taken = \( \require{cancel} \text{2 minute 30 second + 1 minute} = \text{3 minute 30 second}\& = (3×60) +30 \\\\\quad= 180+30 \\\\\quad= 210 \text{ second}\\\\ Speed = \dfrac{Total~Distance}{Total~Time}\\\\\quad\quad=\dfrac{400}{210} \\\\\quad\quad=\dfrac{40\cancel0}{21\cancel0} \\\\\quad\quad= \dfrac{40}{21} m/s \\\\\quad\quad= 1. 90 m/s\\\\ Velocity = \dfrac{Displacement}{Total~time} \\\\\quad\quad= \dfrac{(300-100)}{210} \\\\\quad\quad= \dfrac{200}{210}\\\\\quad\quad= \dfrac{20\cancel{0}}{21\cancel{0}} \\\\\quad\quad= \dfrac{20}{21} m/s \\\\\quad= 0.952m/s \)
fig 7.2
3. Abdul, while driving to school, computes the average speed for his trip to be 20 km h–1. On his return trip along the same route, there is less traffic, and the average speed is 30 km h–1. What is the average speed for Abdul’s trip?
Solution:
Speed of Abdul to school$$=20km/h$$Speed of Abdul to home$$=30km/h$$
Let the distance to school be \(=x~km\)
Time taken to reach school$$=\frac{Distance}{Speed}=\frac{x}{20}$$
Time taken return$$=\frac{x}{30}$$
Total Time of the trip $$\require{cancel}
\\\quad=\frac{x}{2}+\frac{x}{30}\
\\\\\quad=\frac{3x+2x}{60}\
\\\\\quad=\frac{5x}{60}\
\\\\\quad=\frac{\cancel{5}x}{\cancelto{12}{60}}\
\\\\\quad=\frac{x}{12}$$
Total Distance of the trip \(=x+x=2x\)
Average Speed
(\(v_{av}\))$$v_{av}=\dfrac{Total~Distance}{Total~Time}\\\\\quad=\dfrac{2x}{\frac{x}{12}}\\\\\quad=\dfrac{2x\times
12}{x}\
\\\quad=24~km/h$$
4. A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of \(3.0m/ s^2\) for 8.0 s. How far does the boat travel during this time?
Solution:
$$\require{cancel}\text{Initial Velocity }u=0\\
\text{Acceleration }a=3m/s\\
\text{time }t=8s\\
\text{Distance covered }\\s=ut+\dfrac{1}{2}at^{2}\
\\\quad=0\cdot t+\dfrac{1}{2}3\cdot \left( 8\right) ^{2}\
\\\quad=0+\dfrac{1}{2}\times 3\times 8\times 8\
\\\quad=\dfrac{1}{\cancel{2}}\times 3\times 8\times \cancelto{4}{8}\
\\\quad=96m$$
5. A driver of a car travelling at 52 km h–1 applies the brakes. Shade the area on the graph that represents the distance travelled by the car during the period.
(b) Which part of the graph represents uniform motion of the car?
Solution: The Shaded area in Fig. 3 represents the distance travelled by the car when applied
break
(b). There is no information provided on when the graph will represent uniform
motion. We can only assume that before applying the brake, the car was travelling on a uniform
speed. There is also no information on whether the driver continues to apply the brakes.
6. Fig. 7.10 shows the distance-time graph of three objects, A, B, and C. Study the graph and answer the following questions:
- Which of the three is travelling the fastest?
- Are all three ever at the same point on the road?
- How far has C travelled when B passes A?
- How far has B travelled by the time it passes C?
- B is travelling very fast as its slope is steepest.
- No, all three will never meet, as slopes are divergent.
- 7 units = 4 kms, therefore, 1 unit = 4/7 kms. When B passes A, C travels 8 km, but it started at the 4th point, so the distance where C start travelling from is \(\\=\dfrac{4}{7}\times 4 \\\\=
\dfrac{16}{7}\\\)
Total distance travelled by C \(\\\quad=8-\dfrac{16}{7}\\\\\quad=\dfrac{56-16}{7}\\\\\quad=\dfrac{40}{7}\
\\\\\quad=5.71~kms\)
- B travelled to 4.2 points when it passes C, therefore,
Distance travelled by B \(\\4 + 2\times \dfrac{4}{7}\ \\\\\quad=4 + \dfrac{8}{7}\ \\\\\quad=\dfrac{28+8}{7}\ \\\\\quad=\dfrac{36}{7}\ \\\\\quad=5.14~kms \)
7. A ball is gently dropped from a height of 20 m. If its velocity increases uniformly at the rate of \(10m/s^2\), with what velocity will it strike the ground? After what time will it strike the ground?
Solution:
Velocity \(v\) at which the ball strikes the ground
$$\text{Initial Velocity }u=0\\
\text{Acceleration }a=10m/s^{2}\\
\text{Height from ball is dropped }s=20m\\\\
\color{blue}v^{2}-u^{2}=2as\qquad\Rightarrow\text{(Third Equation of Motion)}\\\\
v^{2}-0^{2}=2\times 10\times 20\
v^{2}\\\quad=400\
v\\\quad=\sqrt{40}\
\\\quad=20m$$
Time \(t\) taken to reach ground
$$\begin{aligned}
\text{Initial Velocity }u&=0\\
\text{Final Velocity }v&=20m/s\\
\text{Acceleration }a&=10m/s^{2}\\
\color{blue}v&=\color{blue}u+at\qquad\text{(First Equation of Motion)}\\
20&=0+10t\\
20&=10t\\
\Rightarrow 10t&=20\\
t&=\dfrac{20}{10}\\
&=2~seconds
\end{aligned}$$
8. The speed-time graph for a car is shown in Fig. 7.11.
- Find how far the car travels in the first 4 seconds. Shade the area on the graph that represents the distance travelled by the car during the period.
- Which part of the graph represents uniform motion of the car?
- Let us assume that the parabolic path is nearly straight; then the distance can be calculated as the area under it.\( \require{cancel} \text{Area of the right angled }\Delta =\frac{1}{2}\times base\times height\\\\\quad=\frac{1}{2}\times 4\times 6\\\\\quad=\frac{1}{\cancel{2}}\times \cancelto{2}{4}\times 6\\\\quad=2\times 6\\\\\quad=12m \)
- The graph represents Uniform motion during 6-10 seconds.
fig 7.11
9. State which of the following situations are possible and give an example for each of these:
- an object with a constant acceleration but with zero velocity
- an object moving with an acceleration but with uniform speed.
- an object moving in a certain direction with an acceleration in the perpendicular direction.
- an object with a constant acceleration but with zero velocity: Possible
An object can have zero velocity at a particular instant and still be accelerating. This typically occurs at the turning point of motion.
Example:A ball thrown straight up into the air. At the highest point, the velocity is zero, but it still has a constant downward acceleration due to gravity \(\approx 9.8 m/s^2\). - an object moving with an acceleration but with uniform speed: Conditionally possible: if
it's uniform circular motion
Example: A car moving around a circular track at constant speed.- The speed remains constant.
- The velocity is constantly changing direction.
- The object has centripetal acceleration, pointing towards the centre of the circle.
- an object moving in a certain direction with an acceleration in the perpendicular
direction. Possible
This is exactly what happens in uniform circular motion. The acceleration is perpendicular to the velocity (motion direction) at every point.
Example: A satellite orbiting the Earth in a circular orbit.- It moves tangentially to its orbit (its velocity direction).
- The gravitational force (acceleration) acts towards the centre of the Earth — perpendicular to its motion.
10. An artificial satellite is moving in a circular orbit of radius 42250 km. Calculate its speed if it takes 24 hours to revolve around the Earth.
Solution:
Radius$$r=42250$$Circumference$$\\\quad=2\pi r\\\\\quad=2\times \dfrac{22}{7}\times
42250\
$$Time of Revolution(t) \(=24h\)Speed of Sattelite $$\\\quad=
\frac{Circumference}{time}\\\\\quad=2\cdot \dfrac{22}{7}\times
\dfrac{42250}{24}\\\\\quad=11,065.476km/h\\\\\quad=11,065.476\times\frac{5}{18}\\\\\quad=3,073.74m/s$$