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
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.
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.