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Question
A body moving with a constant acceleration travels distances 3 m and 8 m, respectively in 1 s and 2 s. Calculate:
- The initial velocity.
- The acceleration of body.
Solution
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
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