Question

A body under goes an acceleration of 10m/s^2 in 5 seconds . Find the distance covered by if it its initial velocity is 3m/s

Answer 1

Answer:

The distance covered is 140 meters.

Explanation:

By using the the motion formula $s = ut +\frac{1}{2} at^{2}$, we can say

Distance = 3 × 5 + ( $\frac{1}{2}$ × 10 × 5²)

Distance = 15 + (5 × 25)

Distance = 15 + 125 = 140m

Therefore, the distance covered is 140 meters.

I hope this helps!!

Regards,

Lilac584

Answer 2

We have,

• Initial velocity(u) = 3m/s
• Time taken(t) = 5sec
• Acceleration(a) = 10m/s^2
• Distance covered(s) = ?

We know that,

$\sf\purple{\large{\underline{\underline{s = ut + \frac{1}{2}at^{2} \:_{(2nd \: eq^n \: of \: motion)}}}}}$

$\sf{s = 3(5) + \frac{1}{2} \times10(5)^2}$

$\sf{s = 15 + \frac{1}{2} \times 250}$

$\sf{s = 15 + 125}$

$\sf\purple{\underline{\large\frak{s = 140}}}$

$\sf\pink{\underbrace{\underline{\large{\therefore{Distance \: covered=140m}}}}}$