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A train bracking with constant deceleration covers $1km$ in $20s$, and a second kilometre in $30s$ find the deceleration.
Here is what I did: Since the deceleration is constant it would reach the average velocity in the middle of each $km$ So at $500m$ its velocity is $50m/s$ and at $1500m$ its velocity is $33.3m/s$ Plugging that in the equation$ v^2=u^2 +2as $ we get approximately $0.7m/s$ which is close to the answer but not equal to it.

Can someone tell me where I went wrong and help me answer the question

H_Hassan
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1 Answers1

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While it is true that the average velocity is reached at the half-point in time, it is travelling faster in the first half of the time interval than in the second, so it travels further. Therefore the average velocity over a given 1km will actually happen after the 500m point.

Instead, you would have to observe that given the deceleration is constant, the location is quadratic in the time:

$x_t=v_0t-d\frac{t^2}2$.

Given $x_{20s}=1km$ and $x_{50s}=2km$, solve for $d$.

Brian Jefferson
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