A) \[\frac{\alpha +\beta {{t}^{2}}}{\left( \alpha +\beta \right)}\]
B) \[\frac{\alpha \beta {{t}^{2}}}{2\left( \alpha +\beta \right)}\]
C) \[\frac{{{\alpha }^{2}}t}{\left( \alpha +\beta \right)}\]
D) \[\frac{{{\beta }^{2}}{{t}^{2}}}{2\left( \alpha +\beta \right)}\]
Correct Answer: B
Solution :
| If \[{{t}_{1}}\] is the time of acceleration, then |
| \[\alpha {{t}_{1}}=\beta \left( t-{{t}_{1}} \right)\Rightarrow {{t}_{1}}=\frac{\beta t}{\alpha +\beta }\] |
| \[\therefore \]\[{{v}_{\max }}=\alpha {{t}_{1}}=\frac{\alpha \beta t}{\alpha +\beta },\] |
| \[s=\frac{1}{2}{{v}_{\max }}t.=\frac{\alpha \beta {{t}^{2}}}{2\left( \alpha +\beta \right)}.\] |
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