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.

As we know,

S = ut + $\dfrac{1}{2}$at²

Let,

S₁ = 3 m

t₁ = 1 s

S₂ = 8 m

t₂ = 2 s

Substituting the values in the formula above we get,

$3 = (u \times 1) + (\dfrac{1}{2} \times a \times 1^2) \\[0.5em] 3 = u + (\dfrac{1}{2}a) \\[0.5em] 3 = u + \dfrac{a}{2} \\[0.5em] 2 \times (3 - u) = a \\[0.5em] \Rightarrow a = 6 - 2u \qquad \bold{….[Eqn. \space 1]}$

For 8m distance,

$8 = (u \times 2) + (\dfrac{1}{2} \times a \times 2^2) \\[0.5em] 8 = 2u + (\dfrac{1}{2} \times 4a) \\[0.5em] 8 - 2u = 2a \\[0.5em] 2 \times (4 - u) = 2a \\[0.5em] \Rightarrow a = 4 - u \qquad \bold{….[Eqn. \space 2]}$

Solving Equations 1 & 2,

a = 6 - 2u      [Eqn. 1]

a = 4 - u        [Eqn. 2]

we get,

    0 = 2 - u
⇒ u = 2

Hence, initial velocity = 2 m s ^-1

Putting the value of u in Equation 2 and solving for a:

$a = 4 - u \\[0.5em] \Rightarrow a = 4 - 2 \\[0.5em] \Rightarrow a = 2 m s ^{-2}\\[0.5em]$

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