Question

A train starts from rest at a station and accelerate at a rate of 2 m/s²for 10 s. It then runs at constant speed for 30 s, and slows down at -4 m/s² until it stops at the next station. Find the total distance covered

Answer 1

At the first part of journey the distance can be found from the kinematic equation:

$$d_1 = \dfrac{a_1t_1^2}{2} = \dfrac{2m/s^2\cdot (10s)^2}{2} = 100m$$

where $a_1 = 2m/s^2,t_1 = 10s$.

The speed gained by the end of this part is:

$$v_2 = a_1t_1 = 2m/s^2\cdot 10s = 20m/s$$

At the second part the distance is:

$$d_2 = v_2t_2 = 20m/s\cdot 30s = 600m$$

And the distance for the last part can be found from another kinematic equation:

$$d_3 = \dfrac{v_3^2-v_2^2}{2a_3}$$

where $v_3 = 0m/s$ (train stops), $a_3 = -4m/s^2$. Thus, obtain:

$$d_3 = \dfrac{-(20m/s)^2}{2\cdot (-4m/s^2)} = 50m$$

Finally, the total distance:

$$d = d_1 + d_2 + d_3 = 100m + 600m + 50m = 750m$$

Answer. 750m.