Question

A car having initial velocity 8 m/s moves with a constant acceleration of 8 m/s² on a straight road. The ratio of distance travelled by car in first, third and fifth second of its motion is ?​

Answer 1

Given:

A car having initial velocity 8 m/s moves with a constant acceleration of 8 m/s² on a straight road.

To find:

Ratio of distance travelled by car in first, third and fifth second of its motion?

Calculation:

Distance in 1st sec:

$\rm d_{1} = u + \dfrac{1}{2} a(2t - 1)$

$\rm \implies d_{1} = 8 + \dfrac{1}{2} .8.(2 \times 1 - 1)$

$\rm \implies d_{1} = 8 + 4$

$\rm \implies d_{1} = 12 \: m$

In 3rd sec :

$\rm d_{2} = u + \dfrac{1}{2} a(2t - 1)$

$\rm \implies d_{2} = 8 + \dfrac{1}{2} .8.(2 \times 3 - 1)$

$\rm \implies d_{2} = 8 + 20$

$\rm \implies d_{2} =28 \: m$

In 5th sec:

$\rm d_{3} = u + \dfrac{1}{2} a(2t - 1)$

$\rm \implies d_{3} = 8 + \dfrac{1}{2} .8.(2 \times 5 - 1)$

$\rm \implies d_{3} =44\: m$

So, required ratio :

$= 12 : 28 : 44$

$= 3 : 7 : 11$

Hope It Helps.