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.

answer step by

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

Or uniform acceleration, a = - 1.6

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:

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°