A) \[\frac{L}{v}+\frac{v}{2}\left( \frac{1}{\alpha }+\frac{1}{\beta } \right)\]
B) \[\frac{L}{v}+\frac{2}{v}\left( \frac{1}{\alpha }+\frac{1}{\beta } \right)\]
C) \[\frac{L}{v}+2v\left( \frac{1}{\alpha }+\frac{1}{\beta } \right)\]
D) \[\frac{L}{v}+\frac{1}{v}\left( \frac{1}{\alpha }+\frac{1}{\beta } \right)\]
Correct Answer: A
Solution :
(i) Velocity increases from 0 to v: |
We know that v = u + at Here, u = 0, v = v, \[a=\alpha \therefore t=\frac{v}{\alpha }\] |
and \[s=ut+\frac{1}{2}a{{t}^{2}};{{s}_{1}}=\frac{{{v}^{2}}}{2\alpha }\] |
(ii) Velocity decreases from \[v\] to 0: Using \[v=u+\] at |
Here, \[u=v,v=0,a=-\beta \Rightarrow t=\frac{v}{\beta }\] |
and \[s=ut+\frac{1}{2}a{{t}^{2}};{{s}_{2}}=\frac{{{v}^{2}}}{2\beta }\] |
So, distance travelled during acceleration and retardation, \[d={{s}_{1}}+{{s}_{2}}=\frac{{{v}^{2}}}{2}\left( \frac{1}{\alpha }+\frac{1}{\beta } \right)\] |
(iii) Thus, distance travelled during constant velocity \[=L-\frac{{{v}^{2}}}{2}\left( \frac{1}{\alpha }+\frac{1}{\beta } \right)\] |
So, time taken to travel this distance |
\[T=\frac{L-\frac{{{v}^{2}}}{2}\left( \frac{1}{\alpha }+\frac{1}{\beta } \right)}{v}=\frac{L}{v}-\frac{v}{2}\left( \frac{1}{\alpha }+\frac{1}{\beta } \right)\] |
Hence, total time taken to cover distance L |
\[=\left( \frac{v}{\alpha } \right)+\left[ \frac{L}{v}-\frac{v}{2}\left( \frac{1}{\alpha }+\frac{1}{\beta } \right) \right]+\left( \frac{v}{\beta } \right)\] |
\[=\frac{L}{v}+\frac{v}{2}\left( \frac{1}{\alpha }+\frac{1}{\beta } \right)\] |
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