Question

A body moving with a constant acceleration travels distances 3 m and 8 m, respectively in 1 s and 2 s. Calculate:

1. The initial velocity.
2. The acceleration of body.

Answer 1

Let the constant acceleration with which the body moves be 'a'.

Given, the body travels distance S₁ = 3 m in time t₁ = 1 s.

Same body travels distance S₂ = 8 m in time t₂ = 2 s.

(i) Let 'u' be the initial velocity.

Using the second equation of motion,

S = ut + `1/2` at²

Substituting the values in the formula above we get,

`3 = ("u" xx 1) + (1/2 xx "a" xx 1^2)`

`3 = "u" + (1/2) "a"`

3 = u `+ "a"/2`

2 × (3 - u) = a

⇒ a = 6 - 2u    ....[Equation 1]

For 8 m distance,

`8 = ("u" xx 2) + (1/2 xx "a" xx 2^2)`

`8 = 2"u" + (1/2 xx 4"a")`

8 - 2u = 2a

2(4 - u) = 2a

⇒ a = `(2 (4 - "u"))/2`

⇒ a = 4 - u   ...[Equation 2]

Substracting the Equations 2 from 1

a = 6 - 2u
a = 4 - u
-    -   +  
0 = 2 - u

⇒ u = 2

Hence, initial velocity = 2 m s ^-1

(ii) Putting u = 2 m s in the equation (2)

a = 4 - u

⇒ a = 4 - 2

⇒ a = 2 m s ^-1

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