Question

A train is moving with a velocity of 90 km/h. It is brought to rest by applying the brakes which produce a retardation of 0.5 m/s^2. Find-
1) the velocity after 10 s
2) the time taken by the train to come to rest​

Answer 1

Answer:

hope it helps you............

Answer 2

Answer:

1 km/hr = 1000 meters / 3600 s = 0.2777777… m/s

So 90 km/hr = 90 x 0.277777777.. m/s = 25 m/s

Okay, now the mathematics should be easy. Just use the formula for velocity v and acceleration a over time t:

v = a t

If I am going 25 m/s, and slow down 0.5 m/s² for 10 s, that’s a total of 0.5 x 10 m/s of velocity lost. 5 m/s, to put it simply.

So, if I was going 25 m/s and slowed down by 5 m/s, I’d still be going 20 m/s.

Piece of cake, but let’s convert that back to km/hr just to make the answer fit more neatly with the problem:

20 m/s / 0.2777777777777… m/s/km/hr = 72 km/hr

To come to rest, the train must decelerate to 0 m/s. So, we will just solve our equation for t, instead of solving for v. Rearrange the equation by dividing both sides by a.

t = v / a

25 m/s divided by 0.5 m/s² should tell us the total time it takes to come to rest, then.

t = 25 m/s / 0.5 m/s² = 50 s.