Question

The Speed of train is reduced from 60km/h at the same time as it travels a distance of 450 m. If the retardation is uniform, find how much further it will travel (approximately) before coming to rest?

Answer 1

Answer on Question#53388 - Physics - Mechanics - Kinematics - Dynamics

The Speed of train is reduced from $v = 60\frac{\mathrm{km}}{\mathrm{h}}$ at the same time as it travels a distance of $l = 450\mathrm{m}$ . If the retardation is uniform, find how much further it will travel (approximately) before coming to rest?

Solution:

The time needed for the train moving at the speed $v$ to path the distance $l$ is given by

$$t = \frac {l}{v}$$

Since it took the same time for it to stop, the acceleration is given by

$$a = - \frac {v}{t} = - \frac {v ^ {2}}{l}$$

The distance traveled by the train is given by

$$L = v t + \frac {a t ^ {2}}{2} = v \frac {l}{v} - \frac {v ^ {2} \left(\frac {l}{v}\right) ^ {2}}{2 l} = \frac {l}{2} = 2 2 5 \mathrm {m}$$

Answer: 225 m.