Question

a body moves with an initial velocity ,u and accelerates at the rate , a . show that in time, t . determine the distance covered by the body in 4.0s if it's initial velocity is 20.0ms-1 and the acceleration is 4.0ms-2​

Answer 1

Explanation:

Given,

Time(t) = 4s

Initial Velocity (u) = 20m/s

acceleration (a) = 4m/s²

To Find :-

Distance(S)

Formula Required :-

S = ut + 1/2 at²

Solution :-

S = $\sf 20\times 4 + \dfrac{1}{2}\times 4\times (4)^2$

= $\sf 80 + 2\times 16$

= 80 + 32

= 112m

The Distance Covered by the body (S) = 112m

Some Useful Kinematics Formulas:-

1) S = ut + 1/2 at²

2) S = vt - 1/2 at²

3) a = v - u/t

4) v² - u² = 2as

5) S = (u + v)t/2