Question

A car initially at rest  starts moving with a constant acceleration  of 0.5 m / s ^ 2 and travels a distance of 49m . Find its final velocity and the time taken .​

Answer 1

Given :-

• Car is initially at rest [u =0]
• Acceleration of the body is 0.5 m/s² [a = 0.5 m/s²]
• Distance travelled is 49m [s = 49m]

To find :-

• Final velocity of the body
• Time taken by the body

SOLUTION :-

Firstly lets calculate the time taken by the body through the second equations of motion

$\large\bf\pink{ \ \: s = ut + \dfrac{1}{2} at {}^{2}}$

• s = 49
• u = 0
• a = 0.5

Substituting the values,

$\implies \bf \: 49 = 0(t) + \dfrac{1}{2} (0.5)(t {}^{2} )$

$\implies \bf \: 49 = 0 + \dfrac{1}{2} ( \dfrac{1}{2} )(t {}^{2} )$

$\implies \bf \: 49 = 0 + \dfrac{1}{4}(t {}^{2} )$

$\implies \bf \: 49 = \dfrac{t {}^{2} }{4}$

$\implies \bf \: 49 \times 4= t {}^{2}$

$\implies \bf \: t = \sqrt{49 \times 4}$

$\implies \bf \: t = \sqrt{49 } \times \sqrt{4}$

$\implies \bf \: t = 7 \times 2$

$\implies\bf \: t = 14s$

So, time taken by the body is 14 s

From,

First equations of motion we find the final velocity of the body :-

$\large\bf\pink{\: v = u + at}$

• u = 0
• a = 0.5 m/s²
• t = 14 s

$\bf\implies \: v = 0 + (0.5)(14)$

$\bf\implies \: v = 0 + ( \dfrac{1}{2} )(14)$

$\bf\implies \: v = 0 + 7$

$\bf \implies\: v = 7m/s$

So , final velocity of the body is 7 m/s