Two particles $A$ and $B$ start from rest and move for equal time on a straight line. The particle $A$ has an acceleration $a$ for the first half of the total time and $2 a$ for the second half. The particle $B$ has an acceleration $2 a$ for the first half and $a$ for the second half. Which particle has covered larger distance?

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Solution

For a particle A,
$S_{A 1} = u t + \frac{1}{2} a t^{2}$ here $u = 0, a = a$
$S_{A 1} = \frac{1}{2} \times a \times \frac{T^{2}}{4} = \frac{a T^{2}}{8}$
$S_{A 2} = u t + \frac{1}{2} a_{1} t^{2}$ here $u = a t = a \frac{T}{2}, a = 2 a$
$S_{A 2} = a \frac{T}{2} \times \frac{T}{2} + \frac{1}{2} \times a \times \frac{T^{2}}{4}$
$S_{A 2} = \frac{a T^{2}}{4} + \frac{a T^{2}}{4} = \frac{a T^{2}}{2}$
$S_{A} = \frac{a T^{2}}{2} + \frac{a T^{2}}{8} = \frac{5 a T^{2}}{8}$
For a particle B,similarly
$S_{B 1} = u t + \frac{1}{2} a t^{2}$
$S_{B 1} = 0 + \frac{1}{2} \times 2 a \times \frac{T^{2}}{4} = \frac{a T^{2}}{4}$
$S_{B 2} = u t + \frac{1}{2} a_{1} t^{2}$
$= 2 a \frac{T}{2} \times \frac{T}{2} + \frac{1}{2} \times a \times \frac{T^{2}}{4}$
$= \frac{a T^{2}}{2} + \frac{a T^{2}}{8} = \frac{5 a T^{2}}{8}$
$S_{B} = \frac{5 a T^{2}}{8} + \frac{a T^{2}}{4} = \frac{7 a T^{2}}{8}$
Particle B covered large distance.

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