A body moving with a constant acceleration travels the distances 3 m and 8 m respectively in 1 s and 2 s. Calculate — (i) the initial velocity, and (ii) the acceleration of body.

S = ut + 1/2 at²

Let,

S₁ = 3 m

t₁ = 1 s

S₂ = 8 m

t₂ = 2 s

Substituting the values in the formula above we get,

3 = (u × 1) + 1/2 × a × 1²

or, 3 = u + a × 1/2

or, a × 1/2 = 3 – u

or, a = 2 × (3 – u)

or, a = 6 – 2u —- (i)

For 8m distance,

8 = (u × 2) + 1/2 × a × 2²

or, 8 = 2u + 2a

or, u + a = 4

Put a = 6 – 2u

or, u + 6 – 2u = 4

or, – u = 4 – 6

or, u = 2 ms^-1

Hence, initial velocity = 2 m s ^-1

Putting the value of u in Equation (i)

a = 6 – 2 × 2 = 2 m s^-2

Hence, the acceleration of body is 2 m s^-2