Question

A train accelerating uniformly from rest attains a maximum speed of $40{\text{ms}}^{-1}$ in $20\text{s}$. It travels at this speed for $20\text{s}$ and is brought to rest by uniform retardation in further $40\text{s}$. What is the magnitude average velocity during this period?

• A. $80/3{\text{ms}}^{-1}$
• B. $40{\text{ms}}^{-1}$
• C. $25{\text{ms}}^{-1}$
• D. $30{\text{ms}}^{-1}$

Answer 1

The correct option is C $25{\text{ms}}^{-1}$

Using first equation of motion
$v=u+at$
$40=\left(20\right)a_1(\because u=0)$
$\therefore a_1=2{\text{m/s}}^2$
$\Rightarrow S_1=\frac{1}{2}{a}_1{t}^2$
$\therefore S_1=400\text{m}$
Thereafter it travels with constant speed of $40\text{m/s}$, hence distance travelled will be,
$\Rightarrow S_2=40\times 20=800\text{m}$
Further it travels with uniform retardation until final velocity becomes 0 Putting $t=40\text{s},u=40\text{m/s}$ and $v=0$ in first equation of motion, we get
$0=40+\left(a_2\right)t$
$-40=\left(40\right)a_2$
$\therefore a_2=-1{\text{m/s}}^2$
$\therefore$ retardation = $1{\text{m/s}}^2$
Putting $a=-1{\text{m/s}}^2$ in third equation of motion, we get
$0^2-{40}^2=2(-1)S_3$
$\Rightarrow S_3=800\text{m}$
Now average velocity $=\frac{\text{Total displacement}}{\text{time}}$
$=\frac{400+800+800}{20+20+40}=25\text{m/s}$