Question

The average velocity of a body moving with uniform acceleration after traveling a distance of 3.06m is 0.34m/s. The change in velocity of the body is 0.18m/s. During this time, its acceleration is
A.$0.01m/{{s}^{2}}$
B.$0.02m/{{s}^{2}}$
C.$0.03m/{{s}^{2}}$
D.$0.04m/{{s}^{2}}$

Answer 1

Hint: We are given the average velocity and the change in velocity of the body whose motion is discussed in the question. From these two relations, we could get two linear equations in two variables and by solving them we get the values for initial and final velocity. Now you could simply substitute them along with the given displacement in Newton’s equation to find acceleration.

Formula used:
Expression for average velocity,
$\overline{v}=\dfrac{v+u}{2}$
Newton’s equation of motion,
${{v}^{2}}-{{u}^{2}}=2as$

Complete step by step answer:
In the question are given the average velocity of a body that is accelerated uniformly after traveling 3.06m as 0.34m/s. We are also given the change in velocity of the body as 0.18m/s and we are asked to find the acceleration during this time.
Let us recall the definition of average velocity. Average velocity is basically given by the displacement divided by the time interval in which the displacement occurs. That is,
$\overline{v}=\dfrac{\Delta x}{\Delta t}$
Average velocity over a time interval can also be found by taking the slope of the x-t graph connecting the initial and final positions corresponding to that interval.
Or simply we could get average velocity by taking the average of known initial (u) and final velocities (v) of the body, that is,
$\overline{v}=\dfrac{v+u}{2}$ ………………… (1)
We are given the average velocity as 0.34m/s, therefore (1) becomes,
$\dfrac{v+u}{2}=0.34m{{s}^{-1}}$
$\Rightarrow v+u=0.68m{{s}^{-1}}$ ………………………… (2)
We are given the change in velocity as 0.18m/s. That is,
Final velocity – initial velocity = 0.18m/s
$v-u=0.18m{{s}^{-1}}$ ……………………… (3)
Now let us solve the two linear equations in 2 variables (2) and (3). Adding both equations gives,
$2v=0.86m{{s}^{-1}}$
$\Rightarrow v=0.43m{{s}^{-1}}$ …………………… (4)
From (2),
$u=0.68-v=0.68-0.43$
$\Rightarrow u=0.25m{{s}^{-1}}$ ………………………. (5)
 As per the question,
$s=3.06m$ ……………………….. (6)
Now from Newton’s equations of motion we have,
${{v}^{2}}-{{u}^{2}}=2as$
$\Rightarrow a=\dfrac{{{v}^{2}}-{{u}^{2}}}{2s}$
Substituting (4), (5) and (6) we get,
$a=\dfrac{{{\left( 0.43 \right)}^{2}}-{{\left( 0.25 \right)}^{2}}}{2\times 3.06}$
$\Rightarrow a=\dfrac{0.1849-0.0625}{6.12}$
$\therefore a=\dfrac{0.1224}{6.12}=0.02m{{s}^{-2}}$
Therefore, the acceleration is found to be$0.02m{{s}^{-2}}$and hence the answer is option B.

Note:
Remember that equation (1) is only applicable in the case of uniformly accelerated motion. We know that acceleration is the rate of change of velocity by definition. We could find the time taken by the body for this motion by substituting the known values in another Newton’s equation of motion. That is given by,
$v=u+at$
$\Rightarrow t=\dfrac{v-u}{a}$
$\therefore t=\dfrac{0.43-0.25}{0.02}=9s$
So the time taken for the motion is 9s.