Question

A body starting from rest moves with a
constant acceleration. It moves a distance s1 in first 5 seconds and a distance s2 in next 5 seconds. Prove that Delta s2 = 3s1.​

Answer 1

Answer:

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Answer 2

Given :

A body starting from rest moves with a  constant acceleration .

To Prove :

$s_2=3s_1$

Where , $s_1$ is distance travelled in first 5 seconds and $s_2$ is distance travelled in next 5 seconds .

Solution :

We know , by equation of motion :

$s=ut+\dfrac{at^2}{2}\\\\s=\dfrac{at^2}{2}$

Therefore , distance travelled in first 5 seconds is :

$s_1=\dfrac{a\times 5^2}{2}\\\\s_1=\dfrac{25 a}{2}$

Also , distance travelled in next 5 seconds is :

$s_2=s_{10}-s_1\\\\s_2=\dfrac{10^2a}{2}-\dfrac{25a}{2}\\\\s_2=\dfrac{75a}{2}$

Therefore , equating $s_1$ and  $s_2$ .

We get , $s_2=3s_1$ .

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Kinematics

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