Question

7. An object accelerates from rest to a velocity 27.5 m/s in
10 sec. Then find distance covered by object in next 10
sec
(1) 550 m
(2) 137.5 m
(3) 412.5 m
(4) 275 m​

Answer 1

Answer:

137.5 m

Explanation:

First write down the given data,

so you have a clear picture

v = 27.5 m/s

u = 0(starts from rest)

t = 10 sec

s=?

We know that

$accelaation = \frac{v - u}{t}$

applying that we get that

accelaration = 2.75 m/s²

So now we even have accelaration.

Now there is an equation to find the distance covered

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

so now we have an equation to solve for distance . applying this into our given data

s = 0(as u =0) + (1/2) 2.75 (10)²

s= (1/2)× 2.75×100

275/2 =s

s=137.5 m

Hope you had a good day