Question

A body accelerates uniformly from 10m/s to 25m/s in 5 seconds. Find:
(i) The velocity of the body after the next five seconds.
(ii) The distance covered by the body in that time.

Answer 1

Answer:

after update it's very hard to type such type of questions..

anyway this is your answer mate...

Explanation:

$a = \frac{v - u}{t} \\ = > a = \frac{25 - 10}{5} \\ = > a = \frac{15}{5} \\ = > a = 3m \: per {sec}^{2} \\ = > u = 10m \: per \: sec \\ = > a = 3 \: m \: per {sec}^{2} \\ = > v = u + at \\ = > v = 10 + 3 \times 5 \\ = > v = 25m \: per \: sec \\ = > u = 10m \: per \: sec \\ = > a = 3m \: per \: {sec}^{2} \\ = > t = 5sec \\ = > s = ut + \frac{1}{2} a {t}^{2} \\ = > s = 10 \times 5 + \frac{1}{2 } \times 3 \times {5}^{2} \\ = > s = 50 + \frac{125 }{2} \\ = > s = \frac{100 + 125}{2} \\ = > s = \frac{225}{2} \\ = > s = 112.5 \: m$

Answer 2

Given:

Initial velocity of the body, u = 10 m/s

Final velocity of the body, v = 25 m/s

Time taken = 5 seconds

To Find:

(i) The velocity of the body after next five second.

(ii) The distance covered by the body in that time.

Calculation:

- Applying 1st equation of motion to the given condition:

v = u + at

⇒ 25 = 10 + 5a

⇒ a = 3 m/s²

(i) After the next 5 second, total time becomes:

T = 5 + 5 = 10 seconds

- Final velocity after the next 5 seconds can be calculated by using the 1st equation of motion as follows:

V = u + aT

⇒ V = 10 + (3 × 10)

⇒ V = 40 m/s

(ii) For the next 5 seconds, initial velocity will be:

U = 25 m/s

- Using 3rd equation of  motion, we get:

V² = U² + 2as

⇒ s = (V² - U²) / 2a

⇒ s = (40² - 25²) / (2 × 3)

⇒ s = (1600 - 625) / 6

⇒ s = 975/6

⇒ s = 162.5 m

- So, the answers are as follows:

(i) The velocity of the body after next five seconds is 40 m/s.

(ii) The distance covered by the body in that time is 162.5 m.