- #1

chwala

Gold Member

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- Homework Statement
- A train braking with constant deceleration covers ##1## km in ##20s##, and a second kilometre in ##30s##. Find the deceleration. What further distance will it cover before coming to a stop, and how long will this take?

- Relevant Equations
- Mechanics

Phew; back n forth on this anyway, my lines

...

##v_{1} = \dfrac{1000}{20} = 50## m/s in the first ##10## seconds.

##v_{2} = \dfrac{1000}{30} = \dfrac{100}{3} ##m/s in the first ##35## seconds.

where ##v_1## and ##v_2## are the respective velocities in ##10s## and ##35s## respectively.

Thus,

##a = \dfrac{33.3333 - 50}{35-10} = -\dfrac{2}{3}m/s^2##.

Therefore deceleration is equal to ##\dfrac{2}{3} m/s^2= 0.666m/s^2##.

For next part,

##v=u+at##

##v_{3}= 33.333+(-0.666×15)##

##v_{3} = 23.3333## m/s

using ##v^2 =u^2 +2as##

##0 = 23.333 + (2×-0.666×s)##

##544.3 = 1.333s##

##s = 408.3m## correct to one decimal place.

Finally,

Using ##v= u+at##

##0 = 23.333+ (-0.666t)##

##23.333= 0.66t##

##t =35## seconds

There could be a much better approach hence my post. Cheers.

Last edited: