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Question: A train is travelling at $18\ m/s$. However the conductor notices something on the tracks $126\ m$ ahead of the current position. How long can the conductor wait before beginning to brake if the max safe braking rate is $1.8\ m/s^2$?

I tried this. First I found the braking distance using $a=\frac{v+u}{t}$, then $s=\frac{(v-u)t}{2}$. The distance and time I found was $10\ s$ and $90\ m$. I then subtracted $90$ from $126$ and said he can travel another $36\ m$ before he has to brake. I then found that it would take him $2$ seconds to do so using $v=\frac{s}{t}$ and thus the answer. However, I am unsure. Can someone please tell me if my approach and answers are correct?

zeta
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dreamin
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2 Answers2

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According to Henry's calculation:

$$v=v_0+at=18-1.8 t = 0$$

So $t=10 s$. It will take him 10 sec to safely stop the train.

Opps. I found an error last night. Here is the correct one.

The safe stop distance is then equal to $v_0 t -(1/2)a t^2=180-0.5*1.8*100=90 m$.

Since he spotted the problem with 126 m left and 126-90=36m, 36/18=2 sec. So the driver can wait wait at most 2 sec such that he can safely stop the train.

mike
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This form is more convenient for uniform acceleration: $v^2 - u^2 = 2 a s, u= 0, s = v^2/(2 a) $

It is also obtained by multiplying your two equations.

$ v^2/(2 a)= 18^2/( 2 *1.8) = 90 m $

Maximum distance he can be waiting without taking any action = 126-90 = 36 m

That is 36/18 = 2 seconds, agreeing with all your calculations.

Narasimham
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