A car starts moving rectilinearly, first with acceleration $\alpha=$ $5 \mathrm{~ms}^{-2}$ (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate $\alpha$ comes to a stop. The total time of motion equals $t=25 \mathrm{~s}$. The average velocity…

Question

A car starts moving rectilinearly, first with acceleration $\alpha=$ $5 \mathrm{~ms}^{-2}$ (the initial velocity is equal to zero), then uniformly, and finally, decelerating at the same rate $\alpha$ comes to a stop. The total time of motion equals $t=25 \mathrm{~s}$. The average velocity during this time is equal to $ =72 \mathrm{~km} \mathrm{~h}^{-1}$. How long does the car move uniformly?

Aicharya · Accepted Answer

Let $t$ be the time up to which the car accelerates or decelerates. The maximum velocity attained in this duration is $5 t$. The time $u$ pe which car moves uniformly $=25-2 t$. The velocity-time graph $_{\text {of }}$ the motion of car is drawn as shown in Figure. Given the average velocity is whole time of motion $v_{\mathrm{av}}=\frac{72 \times 5}{18}=20 \mathrm{~ms}^{-1}$ The average velocity from the graph can be obtained as $\begin{array}{ll} v_{\mathrm{av}}=\frac{\text { Total displacement }}{\text { Total time }}=\frac{\text { Area of } \vec{v}-t \text { graph }}{\text { Total time }} \ & 20=\frac{\frac{1}{2} \times[25+(25-2 t)] \times 5 t}{25}=\frac{\left.\frac{1}{2} \times[50-2 t)\right] \times 5 t}{25} \ & 200=50 t-2 t^{2} \ \Rightarrow \quad & t^{2}-25 t+100=0 \ \Rightarrow \quad(t-20)(t-5)=0 \Rightarrow t=5 \mathrm{~s} \text { or } 20 \mathrm{~s} \end{array}$ But $t=20 \mathrm{~s}$ is not possible. Hence, $t=5 \mathrm{~s}$ The time up to which car moves uniformly $=25-2 t=25-2 \times 5=15 \mathrm{~s}$

A car starts moving rectilinearly, first with acceleration \(\alpha=\) \(5 \mathrm{~ms}^{-2}\) (the initial…

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