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?

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)

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?

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.

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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