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Question

A train of length l=350m starts moving rectilinearly with constant acceleration ω=3.0⋅10−2m/s2; t=30s after the start the locomotive headlight is switched on (event 1), and τ=60s after that event the tail signal light is switched on (event 2) . At what constant velocity V (in m/s) relative to the Earth must a certain reference frame K move for the two events to occur in it at the same point? (round off your answer to the nearest integer)

Solution
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In the reference frame fixed to the train, the distance between the two events is obviously equal to l. Suppose the train starts moving at time t=0 in the positive x direction and take the origin (x=0) at the head-light of the train at t=0. Then the coordinate of first event in the earth's frame isx1=12wt2and similarly the coordinate of the second event isx2=12w(t+τ)2−lThe distance between the two events is obviously,x1−x2=l−wτ(t+τ2)=0.242kmin the reference frame fixed on the earth.For the two events to occur at the same point in the reference frame K, moving with constant velocity V relative to the earth, the distance travelled by the frame in the time interval T must be equal to the above distance.Thus Vτ=l−wτ(t+τ2)So, V=1τ−wτ(t+τ2)=4.03m/sThe frame K must clearly be moving in a direction opposite to the train so that if (for example) the origin of the frame coincides with the point x1 on the earth at time t, it coincides with the point x2 at time t+τ.

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