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JEE Mains · Physics · STD 11 - 2. 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

  1. A \(\frac{4 \alpha \beta}{(\alpha+\beta)} t ^{2}\)
  2. B \(\frac{2 \alpha \beta}{(\alpha+\beta)} t ^{2}\)
  3. C \(\frac{\alpha \beta}{2(\alpha+\beta)} t ^{2}\)
  4. D \(\frac{\alpha \beta}{4(\alpha+\beta)} t ^{2}\)
Verified Solution

Answer & Solution

Correct Answer

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

Step-by-step Solution

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…
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