Question

The v-t graph of a moving body is given below. The distance covered by the body in

the first 40s is _________and its acceleration during the last 40s is _____________.


a) 3200m; 2ms-1

b) 700m; -2ms-2

c) 900m; - 2ms-2

d) 900m; 2ms-2​

Answer 1

Answer: The correct answer is a) 3200 m, $-2m/s^2$

Explanation:

Speed is defined as the ratio of distance travelled to the time taken. The equation follows:

$\text{Speed}=\frac{\text{Distance travalled}}{\text{Time taken}}$

From the graph:

Speed for the first 40 s, v = 80 m/s

Time taken, t = 40 s

Putting values in above equation, we get:

$\text{Distance travalled}=(80 m/s\times 40s)=3200 m$

Acceleration is defined as the ratio of change of velocity to the change of time. The equation follows:

$\text{Acceleration}=\frac{\Delta v}{\Delta t}=\frac{v_2-v_1}{t_2-t_1}$

From the graph, for the last 40 sec:

Initial velocity, $v_1$ = 80 m/s

Final velocity, $v_2$ = 0 m/s

Initial time, $t_1$ = 40 s

Final time, $t_2$ = 80 s

Putting values in above equation, we get:

$\text{Acceleration}=\frac{(0-80)m/s}{(80-40)s}\\\\\text{Acceleration}=\frac{-80m/s}{40s}=-2m/s^2$

Hence, the correct answer is a) 3200 m, $-2m/s^2$