Question

A car starts from rest and moves with uniform acceleration a on a straight road from time t = 0 to t = T. After that, a constant deceleration brings it to rest. In this process the average speed of the car is

• A. $\frac{aT}{4}$
• B. $\frac{3aT}{2}$
• C. $\frac{aT}{2}$
• D. $aT$

Answer 1

Answer: (C)

$\frac{aT}{2}$

Answer 2

For First part,

u = 0, t = T and acceleration = a

$\therefore v = 0 + aT = aT$and $S_{1} = 0 + \frac{1}{2}aT^{2} = \frac{1}{2}aT^{2}$

For Second part,

$u = aT,$ retardation =a₁, $v = 0$ and time taken = T₁ (let)

$\therefore 0 = u - a_{1}T_{1} \Rightarrow aT = a_{1}T_{1}$

and from $v^{2} = u^{2} - 2aS_{2}$ $\Rightarrow S_{2} = \frac{u^{2}}{2a_{1}} = \frac{1}{2}\frac{a^{2}T^{2}}{a_{1}}$

$S_{2} = \frac{1}{2}aT \times T_{1}$ $\left( Asa_{1} = \frac{aT}{T_{1}} \right)$

$\therefore$ $v_{av} = \frac{S_{1} + S_{2}}{T + T_{1}} = \frac{\frac{1}{2}aT^{2} + \frac{1}{2}aT \times T_{1}}{T + T_{1}}$

$= \frac{\frac{1}{2}aT\mspace{6mu}(T + T_{1})}{T + T_{1}}$ $= \frac{1}{2}aT$