# 3rd Equation of Motion

## Trending Questions

**Q.**

When a car driver travels at the speed of $10m/s$ applies brake and brings the car to rest in $20sec$. then the retardation will be?

**Q.**A car moving with a speed of 50 km/hr, can be stopped by applying brakes after at least 6 m. If the same car is moving at a speed of 100 km/hr, the minimum stopping distance is

- 6 m
- 12 m
- 24 m
- 18 m

**Q.**A particle starts from rest and has an acceleration of 2 m/s2 for 10 sec. After that, it travels for 30 sec with constant speed and then undergoes a retardation of 4 m/s2 and comes back to rest. The total distance covered by the particle is

- 650 m
- 750 m
- 700 m
- 800 m

**Q.**A car moving with a speed of 40 km/h can be stopped by applying brakes after at least 2 m. If the same car is moving with a speed of 80 km/h, what is the minimum stopping distance

- 4 m
- 6 m
- 8 m
- 2 m

**Q.**A train accelerating uniformly from rest attains a maximum speed of 40 ms−1 in 20 s. It travels at this speed for 20 s and is brought to rest by uniform retardation in further 40 s. What is the magnitude average velocity during this period?

- 80/3 ms−1
- 40 ms−1
- 25 ms−1
- 30 ms−1

**Q.**

Derive the kinematic equations of motion.

**Q.**If the velocity of a particle is given by v=(180−16x)12 m/s then its acceleration will be

- Zero
- 8 m/s
^{2} - – 8 m/s
^{2} - 4 m/s
^{2}

**Q.**If a train travelling at 72 kmph is to be brought to rest in a distance of 200 metres, then its retardation should be

- 20 ms−2
- 10 ms−2
- 2 ms−2
- 1 ms−2

**Q.**A bullet fired into a fixed target loses half of its velocity after penetrating 3 cm. How much further will it penetrate before coming to rest assuming that it faces constant resistance to motion?

- 1.5 cm
- 1.0 cm
- 3.0 cm
- 2.0 cm

**Q.**

Write the equations of motion for freely falling body.

**Q.**

A car moving at 40 km/h is to be stopped by applying brakes in the next 4.0 m. If the car weighs 2000 kg, what average force must be applied on it ?

**Q.**A particle is uncharged and is thrown vertically upward from ground level with a speed of 5√5 ms−1. As a result, it attains a maximum height h. The partide is then given a positive charge +q and reaches the same maximum height h when thrown vertically upward with a speed of 13 ms−1. Finally, the particle is given a negative charge −q. Ignoring air resistance. determine the speed (in ms−1) with which the negatively charged particle must be thrown vertically upward, so that it attains exactly the same maximum height h. (Assume uniform electric field and g=10 ms−2).

**Q.**

A moving train is brought to rest within $20second$ by applying brakes. Find the initial velocity if the relardation due to brakes is $2\mathrm{m}/{\mathrm{s}}^{2}$.

**Q.**A body travels for 15 sec starting from rest with constant acceleration. If it travels distances s1, s2 and s3 in the first five seconds, second five seconds and next five seconds respectively, the relation between s1, s2 and s3 is

- s1=13s2=15s3
- s1=15s2=13s3
- 2s1=13s2=15s3
- s1=16s2=15s3

**Q.**A body moving in a straight line with constant acceleration of 10 ms−2 covers a distance of 40 m in the 4th second. How much distance will it cover in the 6th second?

- 50 m
- 60 m
- 70 m
- 80 m

**Q.**A point moving with constant acceleration from A to B in the straight line AB has velocities u and v at A and B respectively. Find its velocity at C, the mid point of AB.

- √v2−u22
- √v2+u22
- √v2+u24
- √v2−u24

**Q.**A particle starts from rest and travels a distance l with uniform acceleration, then moves uniformly over a further distance 2l and finally comes to rest after moving a further distance 3l under uniform retardation. Assuming entire motion to be rectilinear motion the ratio of average speed over the journey to the maximum speed on its ways is

- 1/5
- 2/5
- 3/5
- 4/5

**Q.**An alpha particle enters a hollow tube of 4 m length with an initial speed of 1kms. It is accelerated in the tube and comes out of it with a speed of 9kms. The time for which it remains inside the tube is

**Q.**Speeds of two identical cars are u and 4u at a specific instant. If the same deceleration is applied on both the cars, the ratio of the respective distances in which the two cars are stopped from that instant is

- 1:1
- 1:4
- 1:8
- 1:16

**Q.**

A person is standing on a weighing machine placed on the floor of an elevator. The elevator starts going up with some acceleration, moves with uniform velocity for a while and finally decelerates to stop. The maximum and the minimum weights recorded are 72 kg and 60 kg. Assuming that the magnitudes of the acceleration and the deceleration are the same, find (a) the true weight of the person and (b) the magnitude of the acceleration. Take g = 9.9 m/s2

**Q.**A ball is released from the top of a tower of height h meters. It takes T seconds to reach the ground. What is the position of the ball in T3 seconds

- h/9 meters from the ground
- 7h/9 meters from the ground
- 8h/9 meters from the ground
- 17h/18 meters from the ground

**Q.**the derivative of function f(x)= log e (2x) w.r.t t is

**Q.**A 150 m long train accelerates uniformly from rest. If the front of the train passes a railway worker 50 m away from the station at a speed of 25 m/s, what will be the speed of the back part of the train as it passes the worker?

- 30 m/s
- 40 m/s
- 50 m/s
- 60 m/s

**Q.**A car is moving with a velocity of 20 m/sec. The driver sees a stationary truck at a distance of 100 m ahead. After some reaction time Δt he applies the brakes, produces a retardation of 4 m/s2. The maximum reaction time to avoid collision will be

- 5 sec
- 2.5 sec
- 4 sec
- 10 sec

**Q.**A driver, having a definite reaction time, is capable of stopping over a distance of 30 m on seeing a red traffic light when speed of the car is 72 km/hr and over a distance of 10 m when the speed is 36 km/hr. Assuming that his reaction time and the acceleration of the car remain the same in all cases, find the distance over which he can stop the car when he is going at 54 km/hr.

- 13.75 m
- 18.75 m
- 15.75 m
- 11.75 m

**Q.**

Find the initial velocity of a train, which is stopped in 20 s by applying brakes. The retardation due to brakes is $1.5m/{s}^{2}$.

**Q.**A man loses 20% of his velocity after running through 108 m. If his retardation is uniform, then the maximum total distance he can cover is

- 300 m
- 192 m
- 218 m
- 324 m

**Q.**When the speed of a car is u, the minimum distance over which it can be stopped is s. If the speed becomes nu, what will be the minimum distance over which it can be stopped during same time? Assume the acceleration is constant and same in both cases.

- sn
- n
- sn2
- n2s

**Q.**A body is dropped from the top of a tower of height 3h. The ratio of the intervals of time to cover the three equal heights h is

- t1:t2:t3=1:3:5
- t1:t2:t3=1:2:5
- t1:t2:t3=√3:√2:1
- t1:t2:t3=1:(√2−1):(√3−√2)

**Q.**A particle is thrown vertically upwards. Its velocity at half of the maximum height is 20 m/s. Then the maximum height attained by it will be: (Take g=10 m/s2)

- 40 m
- 25 m
- 10 m
- 45 m