Question

A body starts from rest and has an acceleration 20cm/sec2. what is the distance coverd by the body in first 8 sec​

Answer 1

Initial velocity = u = 0 m/sec

Acceleration = a = 20 cm/s² = 0.2 m/sec²

Time = t = 8 sec

We need to find distance = d

By the second equation of motion, we have

$d = ut + \frac{1}{2} a {t}^{2}$

$d = (0 \times 8) + \frac{1}{2} \times 0.2 \times 8 \times 8$

$d = 0 + (0.2 \times 4 \times 8)$

$d = 6.4 \: m$

Answer: Distance = 6.4 m

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Answer 2

_______________________________

$\sf\small\orange{Given:-}$

• Acceleration = 20cm/sec^2

• Time = 8 sec

• Initial velocity = 0 m/sec

• Distance travelled = s

$\sf\small\orange{Find:-}$

• Distance covered by the body in first.

$\sf\small\orange{Formula:-}$

•$s = ut + \frac{1}{2} {at}^{2}$

$\sf\small\orange{Solution:-}$

s = ut + 1/2at²

= 0 + 1/2 × 20 × 8 × 8

= 640 cm [ 1 m = 100 cm ]

= 640 /100 m

= 6.4 metre

$\sf\small\orange{Note:-}$

The distance covered by the first is 6.4 meters.

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