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A car starts from rest, moves with an acceleration $a$ and then decelerates at a constant rate $b$ for sometimes to come to rest. If the total time taken is $t$. The maximum velocity of car is given by :
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Verified Answer
The correct answer is:
$\frac{a b t}{(a+b)}$
Let car acceleates for time $t_1$ and decelerates for time $t_2$ then
$t_1+t_2=t$.....(1)
From
v=u+a t
$\begin{aligned} & v =u+a t_1 \\ \Rightarrow v & =a t_1\end{aligned}$
For deceleration
$\begin{aligned} & v=u-a t \\ & 0=a t_1-b t_2 (\because u=v)\\ & a t_1=b t_2 \\ & \Rightarrow \quad t_2=\frac{a t_1}{b} \\ & \therefore \quad t_1+\frac{a t_1}{b}=1 [From Eq. (1)]\\ & \Rightarrow \quad t_1\left(1+\frac{a}{b}\right)=t \\ & \Rightarrow \quad t_1=\frac{b t}{a+b} \\ & \end{aligned}$
$\therefore$ Maximum velocity of car
$v=a t_1=\frac{a b t}{a+b}$
$t_1+t_2=t$.....(1)
From
v=u+a t
$\begin{aligned} & v =u+a t_1 \\ \Rightarrow v & =a t_1\end{aligned}$
For deceleration
$\begin{aligned} & v=u-a t \\ & 0=a t_1-b t_2 (\because u=v)\\ & a t_1=b t_2 \\ & \Rightarrow \quad t_2=\frac{a t_1}{b} \\ & \therefore \quad t_1+\frac{a t_1}{b}=1 [From Eq. (1)]\\ & \Rightarrow \quad t_1\left(1+\frac{a}{b}\right)=t \\ & \Rightarrow \quad t_1=\frac{b t}{a+b} \\ & \end{aligned}$
$\therefore$ Maximum velocity of car
$v=a t_1=\frac{a b t}{a+b}$
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