Question

Find the distance covered by train in 5 sec moving with 10 m/s to 15 m/s and

Answer 1

Answer:

$62.5 \: m.$

Explanation:

$a = \frac{v - u}{t} = \frac{15 - 10}{5} = 1 \: m {s}^{ - 1} \\ \\ {v}^{2} - {u}^{2} = 2as \\ \\ {15}^{2} - {10}^{2} = 2 \times 1 \times s \\ \\ s = \frac{225 - 100}{2} = \frac{125}{2} \\ \\ s = 62.5 \: m.$

Answer 2

Answer:

The required distance covered is 62.5 m

Explanation:

Given :

initial velocity, u = 10 m/s

final velocity, v = 15 m/s

time taken, t = 5 sec

Using v = u + at ,

15 = 10 + a(5)

15 - 10 = 5a

5 = 5a

a = 5/5

a = 1 m/s²

Therefore, the acceleration of the train is 1 m/s²

Now, using  $\bf s=ut+\dfrac{1}{2}at^2$

$\sf s=10(5)+\dfrac{1}{2}(1)(5^2) \\\\ \sf s=50 + \dfrac{25}{2} \\\\ \sf s=50+12.5 \\\\ \sf s=62.5 \ m$

Therefore, the distance covered by the train in 5 sec moving with 10 m/s to 15 m/s is 62.5 m