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Q.
The velocity of train increases uniformly from $20 \,km/h$ to $60 \,km/h$ in $4$ hours. The distance travelled by the train during this period is
AIPMTAIPMT 1994Motion in a Straight Line
Solution:
Initial velocity $(u)=20\, km / h$;
Final velocity $(v)$ $=60\, km / h$ and time $(t)=4$ hours.
velocity $(v)=60=u+a t=20+(a \times 4)$
or, $a=\frac{60-20}{4}=10\, km / h ^{2}$.
Therefore distance travelled in $4$ hours is $s$
$s=u t+\frac{1}{2} a t^{2}=(20 \times 4)+\frac{1}{2} \times 10 \times(4)^{2}=160\, km$.