EASY
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A stone is projected vertically upwards with a velocity of 25 m/s, find the total time taken by the stone to reach the launch point and also calculate the maximum height reached by it.

Important Questions on Motion

EASY
Two bodies of masses ma, mbma> mb are dropped from heights a, b respectively. The ratio of velocities with which they reach ground is:
MEDIUM
A stone is dropped into a well having free water surface at depth 44.1 m. The splash is heard 3.13 sec after the stone is dropped. Find the velocity of sound in air (Take g= 9.8 ms2)
EASY
The initial velocity of a particle is 10 m/s. It is moving with an acceleration of 4 m/s2. The distance covered by the particle after 2 s is
MEDIUM

The velocity – time graph of a moving body is shown in the figure. Which of the following statements is true?

Question Image

MEDIUM
Two steel balls of mass 1 kg and 2 kg and a lead ball of 10 kg are released together from the top of tower 30 m high. Assuming the path to be in vacuum:
EASY
A body is travelling with speed 20 m s-1 having acceleration 4 m s-2 the speed of the body after 2 s is
MEDIUM
A bus of mass 2800 kg is travelling at a speed of 90 kmh. A retarding force of 5600 N is applied on it for a distance of 100 m. Its velocity reduces to
HARD

Three particles A, B and C are thrown from top of a building with same speed. A is thrown upwards, B is thrown downwards and C is thrown horizontally, they hit the ground with speed VA, VB and VC respectively then

EASY

A particle starts its motion from rest under the action of a constant force. If the distance covered in first 10 seconds is S1 and that covered in next 10 seconds is S2 then

MEDIUM
Which of the following is an equation for position - Time relation?
HARD

Derive second equation of motion by graphical method.

MEDIUM

Derive third equation of motion by graphical method.

MEDIUM

Complete the following table.

u (m/s) a (m/s2) t (sec) v=u+at (m/s)
2 4 3 -
- 5 2 20
u (m/s) a (m/s2) t (sec) s=ut+12at2 (m)
5 12 3 -
7 - 4 92
u (m/s) a (m/s2) s (m) v2=u2+2as (m/s)2
4 3 - 8
- 5 8.4 10
MEDIUM
How will the equations of motion for an object moving with a uniform velocity change?
EASY
A motorboat starting from rest on a lake accelerates in a straight line at a constant rate of 3.0 m/s2 for 8.0 s. How far does the boat travel during this time?
EASY

Match the first column with appropriate entries in the second and third columns and remake the table.

S. No. Column 1 Column 2 Column 3
1 Negative acceleration The velocity of the object remains constant A car, initially at rest reaches a velocity of 50 km/hr in 10 seconds
2 Positive acceleration The velocity of the object decreases A vehicle is moving with a velocity of 25 m/s
3 Zero acceleration The velocity of the object increases A vehicle moving with the velocity 10 m/s, stops after 5 seconds
MEDIUM
A car is travelling along the road at 8 m s-1. It accelerates at 1 m s-2 for a distance of 18 m. How fast is it then travelling?
MEDIUM
A car is travelling at  20 m s-1 along a road. A child runs out into the road 50 m ahead and the car driver steps on the brake pedal. What must the car’s deceleration be if the car is to stop just before it reaches the child?
HARD
An object starting from rest travels 20 m in first 2 s and 160 m in next 4 s. What will be the velocity after 7 s from the start?
MEDIUM
A car of mass 1000 kg starts from rest and comes down an inclined slope. If it travels a distance of 300 m in 30 seconds, calculate: (a) acceleration and  (b) force acting on car.