If a body of mass 20 Kg is moving with a velocity of 50 m/s after 10s before it comes to rest the distance covered by a body?
Answers
Answered by
3
m=20kg
u=50m/s
t=10s
v=0
Since v=u+at
So 0=50+a x 10
-50=10a
a= -5m/s^2
So S=ut+1/2at^2
So distance Covered=250m
Hope it is helpful
Answered by
10
Given :
- Initial Velocity (u) = 50 m/s
- Mass of the body (m) = 20 kg
- Time (t) = 10 second
- Final Velocity (v) = 0 m/s
To Find :
- Distance covered by the body (s) = ?
Solution :
First we need to find the Acceleration (a) of the body :
→ a = v - u/t
→ a = 0 - 50/10
→ a = -50/10
→ a = -5 m/s²
By using third kinematical equation of motion we can find the distance covered by the body (s). So, Let's calculate it :
→ v² - u² = 2as
→ (0)² - (50)² = 2 × (-5) × s
→ -2500 = -10s
→ s = (-2500)/-10
→ s = (2500)/10
→ s = 250 m
- Hence,the body covers the distance of 250 m.
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