Question

The average velocity of a body moving with uniform acceleration traveling a distance of $3·06m$ is $0·34m{s}^{-1}$. If the change in velocity of the body is $0·18m{s}^{-1}$ during this time, its uniform acceleration is

• A. $0.01m{s}^{-2}$
• B. $0.02m{s}^{-2}$
• C. $0.03m{s}^{-2}$
• D. $0.04m{s}^{-2}$

Answer 1

The correct option is B

$0.02m{s}^{-2}$

Step1: Given data

Average velocity$\left(V_{avg}\right)=0.34m{s}^{-1}$

Change in velocity$\left(∆v\right)=0·18m{s}^{-1}$

Distance $\left(d\right)=$$3·06m$

Step 2: Formula used

To obtain the acceleration, we will use the first equation of motion

i.e $v=u+at$

Where .$v=$Final velocity, $u=$initial velocity, $a=$acceleration, $t=$time

Step3: Compute the time taken to travel $3·06m$

We know that Time = distance/Speed, So, Putting the values

Here distance = $3.06m$

Speed = $0.34m{s}^{-1}$

$$\begin{gathered} t=\frac{3·06}{0·34} \\ t=9sec \end{gathered}$$

Step 4: Calculating the acceleration

Substitute the given and known values in the following formula,

$\begin{array}{rcl}v& =& u+at\\ v-u& =& at\\ ∆v& =& at\\ 0·18& =& a\times 9\\ a& =& 0·02m{s}^{-2}\end{array}$

So, the uniform acceleration of the body must be $\begin{array}{l}0·02m{s}^{-2}\end{array}$.

Hence, option B is the correct answer.