Question

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 $2a$ for the second half. The particle $B$ has an acceleration $2a$ for the first half and $a$ for the second half. Which particle has covered larger distance?

Answer 1

For a particle A,

$S_{A1}=ut+\frac{1}{2}a{t}^2$ here $u=0,a=a$

$S_{A1}=\frac{1}{2}\times a\times \frac{T^2}{4}=\frac{a{T}^2}{8}$

$S_{A2}=ut+\frac{1}{2}{a}_1{t}^2$ here $u=at=a\frac{T}{2},a=2a$

$S_{A2}=a\frac{T}{2}\times \frac{T}{2}+\frac{1}{2}\times a\times \frac{T^2}{4}$

$S_{A2}=\frac{a{T}^2}{4}+\frac{a{T}^2}{4}=\frac{a{T}^2}{2}$

$S_A=\frac{a{T}^2}{2}+\frac{a{T}^2}{8}=\frac{5a{T}^2}{8}$

For a particle B,similarly

$S_{B1}=ut+\frac{1}{2}a{t}^2$

$S_{B1}=0+\frac{1}{2}\times 2a\times \frac{T^2}{4}=\frac{a{T}^2}{4}$

$S_{B2}=ut+\frac{1}{2}{a}_1{t}^2$

$=2a\frac{T}{2}\times \frac{T}{2}+\frac{1}{2}\times a\times \frac{T^2}{4}$

$=\frac{a{T}^2}{2}+\frac{a{T}^2}{8}=\frac{5a{T}^2}{8}$

$S_B=\frac{5a{T}^2}{8}+\frac{a{T}^2}{4}=\frac{7a{T}^2}{8}$

Particle B covered large distance.