question_answer

A car accelerates from rest at a constant rate \[\alpha \] for some time, after which it decelerates at a constant rate \[\beta \] and comes to rest. When total elapsed time is t, then the maximum velocity acquired by the car is :

A)  \[\frac{({{\alpha }^{2}}+{{\beta }^{2}})t}{\alpha \,\beta }\]

B)  \[\frac{({{\alpha }^{2}}-{{\beta }^{2}})t}{\alpha \,\beta }\]

C)  \[\frac{(\alpha +\beta )t}{\alpha \,\beta }\]

D)  \[\frac{\alpha \beta t}{\alpha \,+\beta }\]

Correct Answer: D

Solution :

Key Idea: When car starts from rest, initial velocity is zero but when car is decelerating, final velocity is zero. From equation of motion, we have \[v=u+\alpha t\] Let \[{{t}_{1}}\] be time of acceleration, and \[{{t}_{2}}\] be time of deceleration. Initial velocity \[u=0\], \[\therefore \] \[v=0+\alpha {{t}_{1}}\] \[{{t}_{1}}=\frac{v}{\alpha }\] ... (i) Similarly, when car is decelerating, the fina velocity is zero, \[\therefore \] \[0=v-\beta \,{{t}_{2}}\] ?. (ii) \[\Rightarrow \] \[{{t}_{2}}=\frac{v}{\beta }\] Total time \[t={{t}_{1}}+{{t}_{2}}\] \[=\frac{v}{\alpha }+\frac{v}{\beta }=v\left( \frac{1}{\alpha }+\frac{1}{\beta } \right)\] \[\Rightarrow \] \[t=v\left( \frac{\alpha +\beta }{\alpha \,\beta } \right)\] \[\Rightarrow \] \[v=\frac{\alpha \beta \,t}{\alpha +\beta }\]