A car moves in a straight line, the car accelerates from rest with a constant acceleration \(\alpha\) on a…

1. \(\frac{s}{v}+v \times\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\)
2. \(s+\frac{v}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\)
3. \(\frac{s}{v}+\frac{s}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\)
4. \(\frac{s}{v}+\frac{v}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\)

Solution

Let the car accelerate for a time \(t_{1}\), move uniformly for a time \(t_{2}\) and then retard for a time \(t_{3}\) as shown in the \(v-t\) graph.


Then the total time \((T)\) of motion of the car is
\(T=t_{1}+t_{2}+t_{3}\) ...(i)
The slope of \(v-t\) graph gives \(\frac{v}{t}=\alpha\) and \(\frac{-v}{t_{3}}=-\beta\)
Then \(t_{1}+t_{3}=\frac{v}{\alpha}+\frac{v}{\beta}\)
...(ii)
Area under \(v-t\) graph gives the total displacement \(s=\) Area of the trapezium \(=\frac{v}{2}\left(t_{2}+T\right)\) This gives \(t_{2}=\frac{2 s}{v}-T\) ...(iii)
Substituting \(t_{1}+t_{3}\) from Eq. (ii) and \(t_{2}\) from Eq. (iii) in Eq.
(i) we have \(T=\frac{v}{\alpha}+\frac{u}{\beta}-\frac{2 s}{v}-T\)
This yields \(T=\frac{s}{v}+\frac{v}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)\) ^

Asked in: JEE Mains - Motion In One Dimension - Test 3

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