Question

What is the distance covered by a train while slowing down from

20 ms−1 to 6 ms−1 with a uniform deceleration of 1.6 ms-2?

Answer Step by Step! The Answer is 113.75, But I m getting -113.75, I don't know how to fix this.

Please Answer Step by Step!!

Answer 1

The distance covered is 113.75 m

Explanation:

As per the question:

The initial velocity of the train, v = 20 m/s

The final velocity of the train, v' = 6 m/s

Uniform deceleration, a = 1.6
`m/s^(2)`

Or uniform acceleration, a = - 1.6
`m/s^(2)`

Here, the body decelerates, i.e., slows down at a uniform rate thus we take acceleration with negative sign.

Now, to find the distance covered, s:

Using the eqn of Kinemetics:

`v'^(2) = v^(2) + 2as`

`6^(2) = 20^(2) + 2(- 1.6)s`

`36 - 400 = - 3.6s`

`s = (- 364)/(- 3.6) = 113.75\ m`