Question

A train starts moving from rest and acquires a velocity of 72 km/h in 5 minutes. If the train moves at
a constant acceleration, calculate the following-
Acceleration
(ü)
Distance travelled by the bus to acquire this velocity.
​

Answer 1

Given

initial velocity of train = u = 0

Acquired velocity = v = 72 km/h = 72 × 5/18 m/s = 20 m/s.

Time taken = 5 minutes = 5 × 60 s = 300 sec.

To find

1. Acceleration of the train
2. Distance travelled by train to acquire this velocity

Calculations

• We will use v = u + at to find out acceleration

=> v = u + at

=> 20 = 0 + 300a

=> 20/300 = a

=> a = 2/30

=> a = 1/15 m/s²

• we will use s = ut + 0.5 at² to find out distance traveled.

s = ut + 0.5at²

=> s = 0.5 × 1/15 × (300)²

=> s = 0.5 × 1/15 × 300 × 300

=> s = 3000 m Ans.

Hence, Acceleration of the train is 1/15 m/s² and distance travelled to acquire velocity of 72 km/h from rest is 3000 m.

Answer 2

Answer:

v = final velocity

u = initial velocity

a = acceleration

t = time

s = distance

u= 0(rest)

v = 72km/h = 72×5/18 = 20 m/s

t = 5min = 5×60 = 300s

1) v = u + at

a = v-u/t

a= 20-0/300

a= 2/30 = 1/15

a= 0.066 m/s²

2) s = ut+1/2at²

s = 0×300 + 1/2×1/15×300×300

s = 3000m

so acceleration of train is 1/15 m/s²

and distance travelled by bus = 3000 m

Hope it helps you :)