Q.1. The displacement of a moving object in a given interval of time is zero. Would the distance travelled by the object also be zero? Justify your answer.
Ans. No, the object might have travelled some distance, provided the initial position and final position of the body coincide.
Example of zero displacement
Q.2. How will the equations of motion for an object moving with a uniform velocity change?
Ans. When an object travels with uniform velocity then v = u and a = 0.
∴ v = u + at ⇒ v = u
► s = ut + (1/2) at2 = ut
► v2 = u2 + 2as ⇒ v2 = u2
Q.3. A girl walks along a straight path to drop a letter in the letterbox and comes back to her initial position. Her displacement-time graph is shown in the figure. Plot a velocity-time graph for the same.
Q.4. A car starts from rest and moves along the jc-axis with constant acceleration of 5 ms-2 for 8 seconds. It then continues with constant velocity. What distance will the car cover in 12 seconds since it started from rest?
Ans: u = 0, a = 5 ms-2, t = 8 s.
= 160 m (for first 8 s)
Velocity at the end of 8 s:
► v = u + at = 0 + 5 x 8
= 40 ms-1, (for last 4 s)
∴ s2 - vt = 40 x 4 = 160 m
= s1 + s2 = 160 m + 160 m
= 320 m
Q.5. A motorcyclist drives from A to B with a uniform speed of 30 km/h-1 and returns back with a speed of 20 km/h-1. Find its average speed.
Let distance travelled from A to B = S km
∴ Total distance = x + x = 2x.
► Time taken to travel from A to B = t1 =
► Time taken to travel from B to A = t2 =
► Total time,
► Average speed =
Q.6. Draw a velocity versus time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height.
Q.7. An object is dropped from rest at a height of 150 m and simultaneously another object is dropped from rest at a height 100 m. What is the difference in their heights after 2 s if both the objects drop with the same acceleration? How does the difference in heights vary with time?
Initially difference between height of two objects:
= 150 - 100 = 50 m
Distance travelled by first obejct in 2 s:
Distance travelled by second object in 2 seconds:
Height of first object at t = 2s:
= 150 - 20 = 130 m
Height of second object at t = 2s:
= 100 - 20 = 80 m
Difference in height after t = 2s:
= 130 - 80 = 50 m
Q.8. An object starting from rest travels 20 m in first 2s and 160 m in next 4s. What will be the velocity after 7s from the start?
Ans. u = 0, s1 = 20 m, s2 = 160 m
u = 0, s = 20 m, t = 2 s
s = ut + at2
or a = 2 s/t2 = 10 ms-2
v = u + at = 10 x 2 = 20 ms-1.
For second interval:
u = 20 ms-1, s = 160 m, t = 4 s
Again, s = ut + at2
160 = 20 x 4 + x a x 16a = 10 ms-2
i.e. 'a' is constant
Velocity after 7 seconds:
v = u + at
= 0 + 10 x 7
= 70 m/s
Q.9. Using following data, draw time-displacement graph for a moving object:
Use this graph to find average velocity for first 4 s, for next 4 s and for last 6 s.
vavg for first 4s = Slope from t = 0s to t = 4s
= =1 m/svavg (for t = 4 s to t = 8 s) is 0 as slope is zero
vavg (for t = 10 s to t = 16 s) == - 1 m/s or the body is moving towards the origin.
Q.10. An electron moving with a velocity of 5 X 104 ms-1 enters into a uniform electric field and acquires a uniform acceleration of 104 ms-2 in the direction of its initial motion.
(a) Calculate the time in which the electron would acquire a velocity double of its initial velocity.
(b) How much distance the electron would cover in this time?
u = 5 x 104 m/s, a = 104 m/s2, v = 2u
(a) v = u + at
► 2u = u + 104t,
► u x 10-4 = t
► 5 x 104 x 10-4 = t
or t = 5 s.
(b) s = ut + at2
= 5 x 104 x 5 + x 104 x (5)2
= 3.75 x 105 m
Q.11. Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between 4th and 5th seconds.
If u = initial velocity, then final velocity = 2m.
For acceleration a:
At t = 4s, s1 = ut + at2 = 4u + 8a
At t = 5s, s2 = ut + at2 = 5u +
S2 - S1 = (5 - 4) u + (25 - 16) = u +
Q.12. Two stones are thrown vertically upwards simultaneously with their initial velocities u1 and u2 respectively. Prove that the ratio of heights reached by them would be in the ratio of u12 : u22 (Assume upward acceleration is -g and downward acceleration to be +g).
► velocities u1 and u2
► heights h1 and h2
► accelerations - g and g
At highest points, velocities v1 = v2 = 0
Thus, v2 = u2 + 2as
⇒ v12 = u12 - 2gh1
and v22 = u22 - 2gh2
⇒ 0 = u12 - 2gh1
and 0 = u22 - 2gh2
⇒ u12 = 2gh1
and u22 = 2gh2
or u12 : u22 = h1 : h2