Question

the speed of a train is reduced from 16.6667 m/sec at the same time as it travels a distance of 450 m if the retardation is uniform, find how much further it will travel before coming to rest?

Answer 1

Answer on Question #48150-Physics-Mechanics-Kinematics-Dynamics

The speed of a train is reduced from $v_{0} = 16.6667\frac{m}{s}$ at the same time as it travels a distance of $s = 450m$ if the retardation is uniform, find how much further it will travel before coming to rest ( $v = 0$ )?

Solution

We have velocity at the initial moment of time $v_{0}$ , then train begin its retardation on distance $s$ to speed $v = 0$ . We need to determine the time $t_{rest}$ needed to stop after train travels distance $s$ .

First we can find acceleration:

$$a = \frac {v ^ {2}}{2 s}.$$

Then we can determine time from retardation begins to stop train:

$$t _ {r e s t} = \frac {v}{a} = \frac {v}{\frac {v ^ {2}}{2 s}} = \frac {2 s}{v} = \frac {2 \cdot 4 5 0}{1 6 . 6 6 6 7} = 5 4 s.$$

Answer: 54 s.

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