Question

A particle starts its motion from rest under the action
of a constant force. If the distance covered in first
10 seconds is S1, and that covered in the first
20 seconds is S2, then [AIPMT (Prelims)-2009]
S2=4S1
Give explanation​

Answer 1

Solution:

Note: We will be using kinematics equation for uniform acceleration type motion.

That is:

$\implies$ $\boxed{\sf{s=ut+\frac{1}{2}at^{2}}}$

Where:

$\implies$ Initial velocity (u) = 0 m/s

$\implies$ (s) = Distance

Now:

Assume the acceleration produced due to the said force be 'A'.

For the first 10 seconds,

$\implies$ $\boxed{\sf{S_{1} =\frac{1}{2} a(10)^{2}}}$

At the end of the 10th second,

The final velocity will be treated as the initial velocity for the $\sf{S_{2}}$ part of journey.

Now:

Calculating final velocity during the first 10 seconds,

$\implies$ $\huge{\boxed{\sf{v=u+at}}}$

That is: v = 10 a

Now,

This 'v' is the initial velocity for $\sf{S_{2}}$ journey.

Hence,

$\implies$ $\sf{S_{2} =10a+\frac{1}{2}a(10)^{2}}$

$\implies$ $\sf{S_{2} =4S_{1}}$

Therefore: $\sf{S_{2} =4S_{1}}$

Answer 2

Answer:

hope it's helpful

Explanation:

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