A car accelerates from rest at a constant rate α for some time after which it decelerates at a constant rate β to come to rest. If the total time elapsed is t seconds, the total distance travelled is

(A) \frac{4 \alpha \beta}{(\alpha+\beta)} t ^{2}

(B) \frac{2 \alpha \beta}{(\alpha+\beta)} t ^{2}

(D) \frac{\alpha \beta}{4(\alpha+\beta)} t ^{2}

JEE Mains · Physics · STD 11 - Motion in straight line

A car accelerates from rest at a constant rate \(\alpha\) for some time after which it decelerates at a constant rate \(\beta\) to come to rest. If the total time elapsed is \(t\) seconds, the total distance travelled is

Pick an option to check your answer.

Verified Solution

Correct Answer

(C) \(\frac{\alpha \beta}{2(\alpha+\beta)} t ^{2}\)

Detailed explanation

\(v _{0}=\alpha t _{1}\) and \(0= v _{0}-\beta t _{2} \Rightarrow v _{0}=\beta t _{2}\)

\(t _{1}+ t _{2}= t\)

\(v _{0}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)= t\)

\(\Rightarrow v _{0}=\frac{\alpha \beta t }{\alpha+\beta}\)

Distance \(=\) area of \(v - t\) graph

\(=\frac{1}{2} \times t \times v _{0}=\frac{1}{2} \times t \times \frac{\alpha \beta t }{\alpha+\beta}=\frac{\alpha \beta t ^{2}}{2(\alpha+\beta)}\)

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