The flabbergasted engineer should do something quickly...

Classical Mechanics Level 4

When a high-speed passenger train traveling at $161 \text{ km/h}$ rounds a bend, the engineer is shocked to see that a locomotive has improperly entered onto the track from a siding and is a distance $D = 676 \text{ m}$ ahead. The locomotive is moving at $29.0 \text{ km/h}$ . The engineer of the high-speed train immediately applies the brakes. What must be the magnitude of the resulting constant deceleration (in $\text{m/s}^{2}$ ) if a collision is to be just avoided?

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The answer is 0.994.

1 solution

Karan Taneja
May 29, 2014

Initial relative speed (u) = 161 - 29 = 132 kmph

Distance to be covered (s) = 676 m (limiting case for just avoiding the collision)

Final relative speed (v) = 0 kmph

From equation of motion,

$v^{2} - u^{2} = 2as$

Converting into appropriate units and solving,

$a = \boxed{-0.994}$

exactly same way..!! =D

Deepanshu Gupta - 6 years, 9 months ago

I was really confused but it turns out it was the units....

A Former Brilliant Member - 6 years, 5 months ago

I really disagree. I believe there is a mistake. In the example above you have decided to take the problem relative to the locomotive, which make u^2 =0. However 's' is no longer the distance when the brakes where first applied but the total distance traveled until the train nearly catches the locomotive. Which is considerable more than then 676m as I will show:

Let: The time for the train to catch the loco is 't', and the distance the train travels be 's', 'a' the acceleration (will be -ve), u the initially train speed (44.7m/s) and v the speed of the train and loco when they nearly touch (8.06m/s).

The distant the train travels is: $676 + vt$

The distance the train travels is also $v^2 = u^2 + a*t/2$

Ed Sirett - 4 years, 8 months ago

[For some reason I can't carry on editting my comment below.....] v^2 = u^2 + 1/2 a t^2. a = (v-u)/t. So eliminating a give t at 105.5s and thus a = -0.35 m/s^2, (also s in total is around 1525m).

Whilst this is a considerable reduction in the deceleration it still exceeds the braking attainable from steel on steel friction.

Ed Sirett - 4 years, 8 months ago

I also got an answer around -0.388 m/s^2. I don't understand why they have to use relative speed here. Why not just plug in the given numbers after unit conversion and be done with it?

Saya Suka - 2 years ago

The flabbergasted engineer should do something quickly...

The answer is 0.994.

1 solution

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