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 |
| \[=\left( \frac{v}{\alpha } \right)+\left[ \frac{L}{v}-\frac{v}{2}\left( \frac{1}{\alpha }+\frac{1}{\beta } \right) \right]+\left( \frac{v}{\beta } \right)\] |
| \[=\frac{L}{v}+\frac{v}{2}\left( \frac{1}{\alpha }+\frac{1}{\beta } \right)\] |
You need to login to perform this action.
You will be redirected in
3 sec
