Question

2) You’re standing on the uOttawa LRT platform watching the trains go by. Two train-cars attached together, each 10 m long, move past you. The first takes 2.10 s to pass you and the second takes 1.65 s. What is the constant acceleration of the train?

Answer 1

Given that length of each train = 10 m

Suppose initial velocity of two-train cars = U m/sec

Given that time taken by train to travel 10 m = t1 = 2.10 sec

Given that time taken by train to travel 20 m = t2 = 2.10 + 1.65 = 3.75 sec

Using 2nd kinematic equation:

d1 = U*t1 + (1/2)*a*t1^2

10 = U*2.10 + (1/2)*a*2.10^2

Also,

d2 = U*t2 + (1/2)*a*t2^2

20 = U*3.75 + (1/2)*a*3.75^2

Now we have two unknown and two equations solving these

2.10*U + 2.205*a = 10

3.75*U + 7.031*a = 20

from first equation:

U = (10 - 2.205*a)/2.10

Using this

3.75*(10 - 2.205*a)/2.10 + 7.031*a = 20

a = 0.693 m/sec^2 = acceleration of train

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