A $$200$$ m long train starts from rest at t $$=$$ $$0$$ with constant acceleration $$4$$ cm $$s-2$$. The head light of its engine is switched on at t $$=$$ $$60$$ s and its tail light is switched on at t $$=$$ $$120$$ s. Find the distance (in$$ $$meter) between these two events for an observer standing on the platform.

Question

Infinity Learn · Accepted Answer

The distance travelled by the head light of engine in 60 s is

$s_{1} = ut + \frac{1}{2} {at}^{2} = 0 \times 60 + \frac{1}{2} \times 0 .04 \times (60)^{2} = 72 m$

The distance travelled by tail light of the train in 120 s is

$s_{2} = ut + \frac{1}{2} {at}^{2} = 0 \times 120 + \frac{1}{2} \times 0 .04 \times (120)^{2} = 288 m$

Thus, the distance between these two events

$= {288 - (200 + 72)} m = 16 m$

Rectilinear Motion

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A 200 m long train starts from rest at t = 0 with constant acceleration 4 cm s^-2. The head light of its engine is switched on at t = 60 s and its tail light is switched on at t = 120 s. Find the distance (in meter) between these two events for an observer standing on the platform.

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Rectilinear Motion

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A 200 m long train starts from rest at t = 0 with constant acceleration 4 cm s-2. The head light of its engine is switched on at t = 60 s and its tail light is switched on at t = 120 s. Find the distance (in meter) between these two events for an observer standing on the platform.

The distance travelled by the head light of engine in 60 s iss1=ut+12at2=0×60+12×0.04×(60)2=72mThe distance travelled by tail light of the train in 120 s iss2=ut+12at2=0×120+12×0.04×(120)2=288mThus, the distance between these two events={288-(200+72)}m=16m

Ready to Test Your Skills?

free study material

A 200 m long train starts from rest at t = 0 with constant acceleration 4 cm s^-2. The head light of its engine is switched on at t = 60 s and its tail light is switched on at t = 120 s. Find the distance (in meter) between these two events for an observer standing on the platform.