The velocity-time graph of a body moving in a straight line is shown in figure.

The ratio of displacement and distance travelled by the body in time $0$ to $10 s$ is

Question

The velocity-time graph of a body moving in a straight line is shown in figure. The ratio of displacement and distance travelled by the body in time 0 to 10 &#160; s is

Aicharya · Accepted Answer

Area of upper rectangles, which will give us magnitude of the area under the velocity-time graph,

$A_{1} = 8 \times 2 + 4 \times 4 = 32$

Area of lower rectangles, which will give us magnitude of the area under the velocity-time graph,

$A_{2} = 4 \times 2 + 4 \times 2 = 16$

Therefore,

$Displacement = A_{1} - A_{2} = 32 - 16 = 16 Distance = A_{1} + A_{2} = 32 + 16 = 48$

Hence, the required ratio would be

$= \frac{16}{48} = \frac{1}{3}$

The velocity-time graph of a body moving in a straight line is shown in figure. The ratio of displacement…

Solution