Question

A train starting from rest attains
a velocity of 72 kmh^-1 in 5
minutes. Assuming that the
acceleration is uniform , find (i) the acceleration, and (ii) the distance travelled by the train for attaining this velocity.
[ANS. (i) 1/15ms^-2 ,(ii) 3.0 km]​

Answer 1

A train starting from rest (initial velocity is 0 m/s) and attains a velocity of 72 kmh^-1 in 5 minutes.

Final velocity of the train is 72 km/hr (72 × 5/18 = 20 m/s) and time is 5 min (5 × 60 = 300 sec).

We have to find the (i) acceleration, and (ii) the distance travelled by the train for attaining this velocity.

(i) For acceleration:

Using the First Equation Of Motion,

v = u + at

Substitute the known in the above formula,

→ 20 = 0 + a(300)

→ 20 = 300a

Divide by 300 on both sides,

→ 20/300 = 300a/300

→ 0.067 = a

Therefore, the acceleration of the train is 0.067 m/s² or 1/15 m/s².

(ii) For distance:

Using the Second Equation Of Motion,

s = ut + 1/2 at²

Substitute the known in the above formula,

→ s = 0(300) + 1/2 × 0.067 × (300)²

→ s = 0 + 0.0335 × 90000

→ s = 0 + 3015

→ s = 3015

Therefore, the distance covered by the train is 3015 m.

To convert m into km. Divide the given value by 100.

So,

The distance covered by the train is 3 km.

Answer 2

ANSWER:

Given

• Train starting from rest that means initial velocity = 0 m/s
• Attains a velocity of 72kmh⁻¹(20m/s) in 5min

To find

1. Acceleration
2. The distance travelled by the train for attaining this velocity

Solution 1:

Time = distance/speed = 300s

Using the kinematic equation

v = u + at

→ v = 0 + a(300)

→ v = 300a

→ 20 = a

→ a = 20/300

→ a = 0.67 m/s² (Answer 1)

________________________

Solution 2:

Using the kinematic equation

s = ut + 1/2 at²

s = 0(t) + 1/2 ( 0.67)(300)²

s = 0 + 0.5 × 0.67 × 90000

s = 3015m

By converting metre to kilometre

s ≈ 3 (Answer 2)

Answers

♪ Acceleration = 0.67 m/s²

♪ Distance travelled by trai. to reach the velocity of 72km/h ≈ 3