Question

a body starts from rest and moves with a constant acceleration it travels a distance S1 in first 10 seconds and and a distance S2 in next 10 seconds find the relation between s2 and as s1​

Answer 1

Answer:

s2 = 10u+s1

Explanation:

Let us consider

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

Initial velocity = 0 I. e., u = 0

s1 =

$s1 = \frac{1}{2} a {10}^{2}$

s1 = 50a

This is first case.

In the second case consider same formula. Acceleration will be same because it is given that acceleration is constant.

Let initial velocity be 'u' here

$s2 = ut + \frac{1}{2} a {t}^{2}$

$s2 = 10u + \frac{1}{2} a {10}^{2}$

$s2 = 10u + 50a$

In first case we got

$s1 = 50a$

now we get

$s2 = 10u + s1$

This is ur answer hope it helps u.

Answer 2

Explanation:

above pic is your answer hope it helps