Question

A body is in motion along a straight line. As it crosses a fixed point a stopwatch is started. The body travels a distance of 1.80m in the first 3 second and 2.20m in the next 5 seconds. What will be the velocity at the end of 9 second?

Answer 1

Using one of Newton’s equations of motion and applying it for the first 3seconds and then for the first 8seconds$(3+5)$you will end up in two linear equations in two variables. Solving them you will get the initial velocity and acceleration of the body. Substituting them and t=9s in another equation of motion, you will get the final velocity at the end of 9 seconds.

Formula used:
Newton’s equations of motion,
$s=ut+\dfrac{1}{2}a{{t}^{2}}$
$v=u+at$

Complete step-by-step answer:
In the question, we are given a body that is in motion along a straight line and a stopwatch is started when this body crosses a fixed point. In the first 3seconds, the body covers a distance of 1.80m and in the next 5seconds the body covers a distance of 2.20m. We are asked to find the velocity of the body at the end of 9 seconds.
In order to solve this, let us recall Newton’s equation of motion which is given by,
$s=ut+\dfrac{1}{2}a{{t}^{2}}$ ……………………………………….. (1)
At the end of 3 seconds,
$s=1.80m$
$t=3s$
So, (1) becomes,
$1.80=3u+\dfrac{1}{2}a{{\left( 3 \right)}^{2}}$
$\Rightarrow 0.6=u+1.5a$ ……………………………….. (2)
After the end of next 5 seconds, that is, at t=8s,
$s=1.80m+2.20m=4m$
Now (1) becomes,
$4=8u+\dfrac{1}{2}a{{\left( 8 \right)}^{2}}$
$\Rightarrow 4=8u+32a$
$\Rightarrow 1=2u+8a$ ………………………………. (3)
(2) ×2will give,
$1.2=2u+3a$………………………….. (4)
Subtracting (3) from (4), we get,
$0.2=-5a$
$a=-0.04m{{s}^{-2}}$
Substituting this in (2),
$0.6=u+\left( 1.5\times -0.04 \right)$
$\Rightarrow u=0.66m{{s}^{-1}}$
We are supposed to find the velocity at the end of 9seconds. So, we have Newton’s equation of motion given by,
$v=u+at$
$\Rightarrow v=0.66-\left( 0.04\times 9 \right)$
$\therefore v=0.3m{{s}^{-1}}$
Hence, we found the velocity at the end of 9seconds to be$0.3m{{s}^{-1}}$.

Note: While solving we have found the acceleration to be negative. Which is a direct implication that the body is decelerating, that is, the velocity of the body is being reduced as the time passes. Also, while substituting in the last part of the solution, we shouldn’t forget the negative sign for acceleration, or else, you will end up answering wrong.