Advertisements
Advertisements
Question
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.

Advertisements
Solution
Velocity-time graph is a straight line parallel to the time axis, so, the velocity of the cyclist is constant.
(i) Acceleration = 0
(ii) At, f = 15 s velocity = 20 m s’1 (from the given graph)
(iii) Distance covered by the cyclist in 15 s
= Area under v - t graph during that time interval
= 20 ms’1 x 15 s
= 300 m.
APPEARS IN
RELATED QUESTIONS
What is meant by the term acceleration ? State the SI unit of acceleration.
Derive the formula s= `ut+1/2at^2` , where the symbols have usual meanings.
A freely falling object travels 4.9 m in 1st second, 14.7 m in 2 nd second, 24.5 m in 3rd second, and so on. This data shows that the motion of a freely falling object is a case of :
Define velocity. State its unit.
Give one example of following motion :
Variable acceleration
A car starting from rest acquires a velocity 180m s-1 in 0.05 h. Find the acceleration.
The figure shows the displacement - time graph for four bodies A, B C and D. In each case state what information do you get about the acceleration (zero, positive or negative).

The velocity-time graph of a body in motion is a straight line inclined to the time axis. The correct statement is ___________
Write three equations of uniformly accelerated motion relating the initial velocity (u), final velocity (v), time (t), acceleration (a) and displacement (S).
A packet is dropped from a stationary helicopter, hovering at a height ‘h’ from ground level, reaches the ground in 12s. Calculate
- the value of h
- final velocity of packet on reaching the ground. (Take g = 9.8 ms−2)
