Question

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

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:

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

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

$\sf \implies \: a = \dfrac{576}{112}$

$\sf\implies \: a = 5.14 \: m {s}^{ - 2}$