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
- A \(\frac{4 \alpha \beta}{(\alpha+\beta)} t ^{2}\)
- B \(\frac{2 \alpha \beta}{(\alpha+\beta)} t ^{2}\)
- C \(\frac{\alpha \beta}{2(\alpha+\beta)} t ^{2}\)
- D \(\frac{\alpha \beta}{4(\alpha+\beta)} t ^{2}\)
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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