Physics, asked by vaishnavi150307, 9 months ago

A body accelerates uniformly from 10 m/s to 25 m/s in 5 s. Find (a) The acceleration (b) The distance covered by the body in that time

Answers

Answered by belalrahman
9

ANSWERS:

Given:

Initial velocity, u = 10 m/s

Final velocity, v = 25 m/s

Time = 5 s

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:

v = u + at

⇒ 25 = 10 + 5a

⇒ a = 3 m/s²

(i) After next 5 second, T = 5 + 5 = 10 s

- Again using 1st equation of motion:

V = u + at

⇒ V = 10 + (3 × 10)

⇒ V = 40 m/s

(ii) For the next 5 seconds, 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,

(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.

Answered by Anonymous
14

A body accelerates uniformly from 10 m/s to 25 m/s in 5 s.

From above data we have the initial velocity of the body i.e. u is 10 m/s, final velocity i.e. v is 25 m/s and time 't' is 5 sec.

We have to find the acceleration and the distance covered by the body in that time.

a) For Acceleration:

Using the First Equation Of Motion,

v = u + at

Substitute the known values,

→ 25 = 10 + a(5)

→ 25 - 10 = 5a

→ 15 = 5a

→ 3 = a

Therefore, the acceleration of the body is 3 m/s².

b) For Distance:

Using the Second Equation Of Motion,

s = ut + 1/2 at²

Substitute the values,

→ s = 10(5) + 1/2 × 3 × (5)²

→ s = 50 + 3/2 × 25

→ s = 50 + 37.5

→ s = 87.5

Therefore, the distance covered by the body is 87.5 m.

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