Question

You’re standing on the 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

$$d=ut_1+0.5at_1^2,d=(u+at_1)t_2+0.5at_2^2$$

$$u=\frac{d}{t_1}-0.5at_1$$

$$d=\left(\frac{d}{t_1}-0.5at_1+at_1\right)t_2+0.5at_2^2=\left(\frac{d}{t_1}+0.5at_1\right)t_2+0.5at_2^2$$

$$10=\left(\frac{10}{2.1}+0.5a(2.1)\right)1.65+0.5a(1.65)^2$$

$$a=0.693\frac{m}{s^2}$$