Question

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

• A. $80/3m{s}^{-1}$
• B. $40m{s}^{-1}$
• C. $25m{s}^{-1}$
• D. $30m{s}^{-1}$

Answer 1

The correct option is C $25m{s}^{-1}$

$\therefore 40=\left(20\right)a_1$
$\therefore a_1=2m/s^2$
Further $40=\left(40\right)a_2$
$\therefore a_2=1m/s^2$
Therefore acceleration is $2m/s^2$ and retardation $1m/s^2$
Now, $S_1=\frac{1}{2}{a}_1{t}_1^2=\frac{1}{2}\times 2\times (20{)}^2=400m$
$S_2=v_{max}{t}_2=40\times 20=800m$
$S_3=\frac{V_{max}^2}{2a_2}=\frac{(40{)}^2}{2\times 1}=800m$
Now average velocity
$=\frac{\text{Total displacement}}{\text{Total time}}$
$=\frac{400+800+800}{20+20+40}=25m/s$.