Motion NCERT Solution, study material, CBSE Notes

NCERT Solution: Motion

How will the equations of motion for an object moving with a uniform velocity change?

When the object is moving with a uniform velocity, then ν = � and a = 0. In this situation, equation for distance would be as follows:

s = ut and ν² - �² = 0

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 Fig. 8.4. Plot a velocity–time graph for the same.

Ans. For the initial 50 second, velocity is 2 m/s. After that, velocity drops of zero; as shown by vertical line in graph.

For the next 50 second, velocity is taken in negative because displacement is becoming zero.

A car starts from rest and moves along the x-axis with constant acceleration 5 ms^–2 for 8 seconds. If it then continues with constant velocity, what distance will the car cover in 12 seconds since it started from the rest?

The distance travelled in first 8 s, x₁ = 0 + 1/2 (5)(8)² =160 m
At this point the velocity v = u + at= 0 + (5�8) = 40 m s^�1
Therefore, the distance covered in last four seconds, x₂ = (40�4) m = 160 m
Thus, the total distance x = x₁ + x₂ = + = (160 + 160) m = 320 m

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 AB = x, So t₁ = x /30 and t₂ = x/20

Total Time = t₁ + t₂ = 5x / 60 hr

Average speed for entire journey = Total distance / Total Time

2x /5x /60 = 24 km hr^-1

The velocity-time graph (Fig. 8.5) shows the motion of a cyclist. Find (i) its acceleration (ii) its velocity and (iii) the distance covered by the cyclist in 15 seconds.

Ans. (i) Since velocity is not changing, acceleration is equal to zero.
(ii) Reading the graph, velocity = 20 ms-1
(iii) Distance covered in 15 seconds, s= u � t= 20 � 15 = 300 m

Draw a velocity versus time graph of a stone thrown vertically upwards and then coming downwards after attaining the maximum height.

When a stone is thrown upward, its velocity is at maximum. The velocity begins to drops as the stone attains height. Once the stone attains the maximum height, velocity becomes zero. After that, stone begins to fall down. At this points, velocity beings to rise. Velocity is at its maximum when the stone hits the ground.

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 same accelerations? How does the difference in heights vary with time?

Initial difference in height = (150�100) m = 50 m