Question

6. A 21,000 kg train is traveling at 21m/s. It accelerates as it enters a tunnel because of a constant force of 45,000 N from the engine. The train is in the tunnel for 5.25 seconds. What is the final speed of the train?

Answer 1

m= 21000 kg

u=21m/s

f=45000N

t=5.25 s

speed = distance / time

s = s vector (because of straight

f=ma

f/m=a

45000/21000=a=2.5m/s²

s=ut+1/2at²=21(5.25)+1/2(2.5)(5.25)²

110+ 69=179

speed=179/5.25

Answer 2

The final speed of the train is 42 m/s.

Explanation:

Given mass of train, m = 21,000 Kg

Initial speed of train, v = 21 m/s

Force from the engine, F = 45,000 N

Train in tunnel for time, t = 5.25 s.

So as per newton`s second law,

F = ma

Here, a is acceleration of train.

Substitute the values, we get

45,000 N = 21,000 Kg (a)

a = 2.14 $m/s^2$

Now to calculate the final speed of train use equation of motion as,

v = u +at

Here, v is the final speed and u is initial speed.

Substitute the values, we get

v = 21 +(9.8) (2.14)      

v = 42 m/s.

Thus, the final speed of the train is 42 m/s.

Topic: Equation of motion

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