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

Answer:

The distance covered is 113.75 m

Step-by-step 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:

$v'^{2} = v^{2} + 2as$

$6^{2} = 20^{2} + 2(- 1.6)s$

$36 - 400 = - 3.6s$

$s = \frac{- 364}{- 3.6} = 113.75\ m$