Question

A train accelerates from 36 km h to 72 km h on covering a distance of 50 m .Calculate the acceleration of the train and the time taken to cover the distance

Answer 1

u=36km/h=10m/s
v=72km/h=20m/s
s=50m
v^2-u^2=2as
20^2-10^2=2×a×50
400-100=2×a×50
a=300/100
a=3m/s^2
time taken to cover the distance is
a=v-u/t
t=v-u/a
t=20-10/3
t=10/3
t=3.3s

Answer 2

Given :

Train accelerates from 36 km/h to 72 km/h on covering a distance of 50 km.

To Find :

The acceleration of the train and the time taken to cover the distance .

Solution :

Acceleration is given by :

$v^2-u^2=2as\\\\a=\dfrac{72^2-36^2}{2\times 50}\\\\a=38.88\ m/s^2$

Time taken is given by :

$v=u+at\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{72-36}{38.88}\\\\t=0.93\ hr$

Hence , this is the required solution .

Learn More :

Equation of motion

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