Question

A train starts from rest and accelerates uniformly at a rate of $2m{s}^{-2}$ for $10s$. It then maintains a constant speed for $200s$. The brakes are then applied and the train is uniformly retarded and comes to rest in $50s$. Find the average velocity of the train.

• A. $35.4m{s}^{-1}$
• B. $20m{s}^{-1}$
• C. $17.69m{s}^{-1}$
• D. $8m{s}^{-1}$

Answer 1

The correct option is C $17.69m{s}^{-1}$

Total distance covered($s$) = Distance during acceleration($s_1$) + distance during uniform motion($s_2$) + distance during retardation($s_3$)

For acceleration,

$v=u+a\ast t$

$v_{max}=2\times 10=20m/s$

$s=ut+\left(1/2\right)a{t}^2$

$s_1=\left(1/2\right)\times 2\times {10}^2$

$=100m$

For uniform motion,

$s=vt$

$s_2=20\times 200$

$=4000m$

During retardation,

$v=u+at$

$0=20+50t$

$=-0.4m/s^2$

$s_3=ut+\left(1/2\right)a{t}^2$

$=20\times 50+\left(1/2\right)\times (-0.4)\times {50}^2$

$=500m$

Total distance travelled, $s=s1+s2+s3=100+4000+500=4600m$

Total time taken, $t=t_1+t_2+t_3=260s$

Average velocity, $v_{avg}=\frac{\text{Total Distance}}{\text{Total Time}}$

$=4600/260=17.69m/s$