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Let initial velocity of the two train cars is u m/s Time taken by the first train to travel 10 m () 2.10 s Time taken by the second train to travel 20 m(t,)(2.10+1.65)s -3.75 s Use the second equation of motion for the first train-car as, at Xut 2 Substitute 10 m for x and 2.10 sfortin the above equation 1 10 m u(2.10 s+a(2.10 s) (1) Use the second equation of motion for the second train-car as, 1 2 X2utat, Substitute 20 m for xand 3.75 sfor t, in the above equation 20 m -u(3.75 s)+a(3.75 s (2) Solve equation (1) and (2) for u and a 10 2.10 u2.205 a (3) (4) 20 3.75 u 7.031 a
10-2.205 a for u in the equation (4) Substitute 2.10 20 3 7510-2.205 a+7.031 a 2.10 a 0.693 m/s moving with a constant acceleration of 0.693 m/s2. Thus, the train is
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