Question

49.An object accelerates from rest to a speed of 24 m/s over a distance of 56 m. What

acceleration did it experience?

Answer 1

Given:

Initial velocity,u = 0 m/s

Final velocity,v = 24 m/s

Distance covered,s = 56 m

To be calculated:-

Calculate the acceleration ( a ) .

Solution:-

According to the third equation of motion,

`\bf \: {v}^(2) = {u}^(2) + 2as`

★ Substituting the values in the third equation of motion:

`\sf \implies \: {(24)}^(2) = {(0)}^(2) + 2 * a * 56`

`\sf \implies \: 576 = 112a`

`\sf \implies \: a = (576)/(112)`

`\sf\implies \: a = 5.14 \: m {s}^( - 2)`