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 :
Let car accelerates for time \[{{t}_{1}}\] and decelerates for time \[{{t}_{2}}\]then \[{{t}_{1}}+{{t}_{2}}=t\] ?(1) From \[v=u+at\] \[v=u+a{{t}_{1}}\] \[\Rightarrow \] \[v=a{{t}_{1}}\] For deceleration \[v=u-at\] \[0=a{{t}_{1}}-b{{t}_{2}}\] \[(\because u=v)\] \[a{{t}_{1}}=b{{t}_{2}}\] \[\Rightarrow \] \[{{t}_{2}}=\frac{a{{t}_{1}}}{b}\] \[\therefore \] \[{{t}_{1}}+\frac{a{{t}_{1}}}{b}=1\] [From Eq. (1)] \[\Rightarrow \] \[{{t}_{1}}\left( 1+\frac{a}{b} \right)=t\] \[{{t}_{1}}=\frac{bt}{a+b}\] \[\therefore \] Maximum velocity of car \[v=a{{t}_{1}}=\frac{abt}{a+b}\]
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