Question

A 200 m long train starts from rest at t-0 with constant acceleration 4 cm/s2.the head light of its engine is switched on at t-60 s and its tail light is switched on at t-120 s.the distance between these two events for an observer standing on platform​

Answer 1

ANSWER:

• The distance between the two events for an observer standing on platform = 216 m.

GIVEN:

• A 200 m long train starts from rest at t = 0 with constant acceleration 4 cm/s².

• The head light of its engine is switched on at t = 60 s and its tail light is switched on at t = 120 s.

TO FIND:

• The distance between the two events for an observer standing on platform.

EXPLANATION:

$\boxed{\bold{\large{\gray{S = ut + \frac{1}{2} at^2}}}}$

Let S₁ be the distance travelled in 60 s.

$\sf S_1 = ut_1 + \dfrac{1}{2} at_1^2$

u = 0

t₁ = 60 s

a = 4 cm/s² = 0.04 m/s²

$\sf S_1 = 0(t_1) + \dfrac{1}{2} 0.04(60)^2$

$\sf S_1 = 0.02(3600)$

$\sf S_1 =2(36)$

$\sf S_1 =72 \ m$

Let S₂ be the distance travelled in 120 s.

$\sf S_2 = ut_2 + \dfrac{1}{2} at_2^2$

u = 0

t₂ = 120 s

a = 4 cm/s² = 0.04 m/s²

$\sf S_2 = 0(t_2)+ \dfrac{1}{2} 0.04(120)^2$

$\sf S_2 = 0.02(14400)$

$\sf S_2 = 2(144)$

$\sf S_2 =288 \ m$

The distance between the two events for an observer standing on platform = S₂ - S₁

S₂ - S₁ = 288 - 72

S₂ - S₁ = 216 m.

Hence the distance between the two events for an observer standing on platform = 216 m.