The velocity-time graph for a particle moving | vedclass

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Question

(B) The average velocity is defined as the total displacement divided by the total time interval: $v_{avg} = \frac{\Delta x}{\Delta t}$.Displacement $

Vedclass · Accepted Answer

$12.5$

Vedclass · Answer

(B) The average velocity is defined as the total displacement divided by the total time interval: $v_{avg} = \frac{\Delta x}{\Delta t}$.Displacement $\Delta x$ is the area under the velocity-time graph between $t=4 \,s$ and $t=6 \,s$.At $t=6 \,s$, $v=15 \,ms^{-1}$. The graph is a straight line from $(0,0)$ to $(6,15)$, so the equation for velocity is $v(t) = \frac{15}{6}t = 2.5t$.At $t=4 \,s$, $v(4) = 2.5 	imes 4 = 10 \,ms^{-1}$.The area under the graph between $t=4 \,s$ and $t=6 \,s$ is a trapezoid with parallel sides $v(4)=10 \,ms^{-1}$ and $v(6)=15 \,ms^{-1}$, and height $\Delta t = 6-4 = 2 \,s$.Area = $\frac{1}{2} 	imes (v(4) + v(6)) 	imes \Delta t = \frac{1}{2} 	imes (10 + 15) 	imes 2 = 25 \,m$.Average velocity $v_{avg} = \frac{25 \,m}{2 \,s} = 12.5 \,ms^{-1}$.Thus, the correct option is $B$.

The velocity-time graph for a particle moving in a straight line is shown in the figure. The average velocity between $t=4 \,s$ and $t=6 \,s$ is . . . . . . (in $\,ms^{-1}$)

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