Question

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?

Answer 1

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

$s = 50 \times 10 + \frac{1 }{2} \times ( - 5) \times {10}^{2} \\ s = 500 - 5 \times 50 \\ s = 500 - 250 \\ = 250$

So distance Covered=250m

Hope it is helpful

Answer 2

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.