Question

A train starting from rest attains a velocity of 72km/h in 5minutes . Assuming that the acceleration is uniform, - the acceleration , the distance travelled by the train for attaining this velocity

Answer 1

Initial velocity (u)=0
Final velocity (v)=72km/h
=20m/s
Time (t)= 5 minute
=5×60 =300s

Acceleration (a)=(v-u)/t
=20/300
=1/15

So, the acceleration is 1/15.

Let distance travelled be s

Using second equation of motion
ut+(1/2)at^2
=(0)(300)+ (1/2)(1/15)(300^2)
= 3000


So the body covered 3000m (3km).

Answer 2

Given :

• u = 0
• v = 72 km/h
• t = 5 min

To find :

• the acceleration
• the distance travelled by the train for attaining this velocity

Solution :

u = 0

v = 72 km/h = (72 x 100 m)/(60 x 60 s) = 20 m/s

t = 5 min = 5 x 60 = 300 s

thus,

a = (v - u)/t

= (20 - 0)/300

= 1/15 m/s²

from,

v² - u² = 2 as

s = (v² - u²)/ 2a

= (20² - 0²)/2 x (1 / 15)

= 3000 m

= 3 km

=> 1/15 m/s² is the acceleration and 3 km distance travelled by the train for attaining this velocity