The velocity-time graph of a body moving alon | vedclass

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(A) The average velocity is given by the total displacement divided by the total time. The displacement is equal to the area under the velocity-time g

Vedclass · Accepted Answer

$1 : 1$

Vedclass · Answer

(A) The average velocity is given by the total displacement divided by the total time. The displacement is equal to the area under the velocity-time graph.Let $V_{max}$ be the maximum velocity at time $t_1$. From the graph,we have:$V_{max} = t_1 	an(30^\circ) = t_2 	an(60^\circ)$$V_{max} = t_1 \left(\frac{1}{\sqrt{3}}ight) = t_2 (\sqrt{3})$Therefore,$\frac{t_1}{t_2} = \frac{\sqrt{3}}{1/\sqrt{3}} = 3$,which means $t_1 = 3t_2$.The displacement during time $t_1$ is $S_1 = \frac{1}{2} 	imes t_1 	imes V_{max}$.The average velocity during $t_1$ is $V_{avg1} = \frac{S_1}{t_1} = \frac{1}{2} V_{max}$.The displacement during time $t_2$ is $S_2 = \frac{1}{2} 	imes t_2 	imes V_{max}$.The average velocity during $t_2$ is $V_{avg2} = \frac{S_2}{t_2} = \frac{1}{2} V_{max}$.Thus,the ratio of the average velocities is $\frac{V_{avg1}}{V_{avg2}} = \frac{\frac{1}{2} V_{max}}{\frac{1}{2} V_{max}} = 1 : 1$.

The velocity-time graph of a body moving along a straight line is shown in the figure. The ratio of the average velocities during the time intervals $t_1$ and $t_2$ is:

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