question_answer

A car starts from rest, moves with an acceleration a and then decelerates at a constant rate b for some time to come to rest. If the total time taken is t. The maximum velocity of car is given by:

A)  \[\frac{abt}{(a+b)}\]                     

B)  \[\frac{{{a}^{2}}t}{a+b}\]

C)   \[\frac{at}{(a+b)}\]                      

D)  \[\frac{{{b}^{2}}t}{a+b}\]

Correct Answer: A

Solution :

                 From equation of motion, we have \[v=u+at\] Since, body starts from rest\[u=0\]. Let\[{{t}_{1}}\]be time when body accelerates and\[{{t}_{2}}\]when it decelerates. \[\therefore \]                  \[t={{t}_{1}}+{{t}_{2}}\] \[\Rightarrow \]                               \[{{t}_{2}}=t-{{t}_{1}}\] \[\therefore \]                  \[v=0+a{{t}_{1}}\]                        ...(1) When, car finally comes to rest\[v=0\] \[\therefore \]                  \[0=v-b(t-{{t}_{1}})\]                  ...(2) From Eqs. (1) and (2), we get                 \[{{t}_{1}}=\frac{b}{a+b}t,v=\frac{ab}{a+b}t\]