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Mechanics help

An engine and train weigh 250 tonnes. The engine exerts a pull of 150000. The resistance of motion is 1/100 of the weight of the train and it's braking force is 1/10 of the weight.
The train starts from rest and accelerates uniformly until it reaches a speed of 45 km/h. At this point the brakes are applied until the train stops.
Find the time taken for the train to stop to the nearest second.
Answer = 36.4


What I did :

F=ma
The resistance of motion=
1/100 × 250 000
= 2500

Braking force=
1/10 × 250 000
= 25000

Resultant force when car decelerates =
(2500 +25000)-150000
=-122500

F=ma
-122500=250000 a
a=-0.49

45 km/h = 12.5 m/s
t= (v-u) /a
= - 12.5/-0.49
=25.5

What did I do wrong?
(edited 6 months ago)
Reply 1
Original post by raghadgamil
An engine and train weigh 250 tonnes. The engine exerts a pull of 150000. The resistance of motion is 1/100 of the weight of the train and it's braking force is 1/10 of the weight.
The train starts from rest and accelerates uniformly until it reaches a speed of 45 km/h. At this point the brakes are applied until the train stops.
Find the time taken for the train to stop to the nearest second.
Answer = 36.4


What I did :

F=ma
The resistance of motion=
1/100 × 250 000
= 2500

Braking force=
1/10 × 250 000
= 25000

Resultant force when car decelerates =
(2500 +25000)-150000
=-122500

F=ma
-122500=250000 a
a=-0.49

45 km/h = 12.5 m/s
t= (v-u) /a
= - 12.5/-0.49
=25.5

What did I do wrong?

A few things.

You need to calculate the time for both the acceleration and the braking phases.

The mass is 250 000 kg and the weight is 250 000g N.

In the acceleration phase you have driving and resistance forces

In the braking phase you have braking and resistance forces.

Reply 2
Original post by mqb2766
A few things.

You need to calculate the time for both the acceleration and the braking phases.

The mass is 250 000 kg and the weight is 250 000g N.

In the acceleration phase you have driving and resistance forces

In the braking phase you have braking and resistance forces.


oh okay ill try again.

When accelerating :
F=ma
150000-25000 = 250 000 a
a = 0.5
t = 12.5/0.5
=25

When decelerating :
F=ma
250 000 +25000 = 250 000 a
a = -1.1
t= -12.5/-1.1
=11.36

total time = 36.4

Thanks! ( :
(edited 5 months ago)

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