Question

The position of the particle is given by $x=2t^2-4t+4$ in $\text{m}$. The average speed of the particle in $6\text{sec}$ is

• A. $\frac{43}{6}\text{m/s}$
• B. $\frac{26}{3}\text{m/s}$
• C. $\frac{27}{3}\text{m/s}$
• D. $\frac{23}{3}\text{m/s}$

Answer 1

The correct option is B $\frac{26}{3}\text{m/s}$

As we know that the velocity of the particle as a function of time is given by
$\frac{dx}{dt}=\frac{d(2t^2-4t+4)}{dt}$
$\Rightarrow v=4t-4$
Now, $v\left(t\right)=4t-4$, we will check if the particle has changed its direction or not within the asked time interval, i.e. the time when velocity of the particle becomes zero.
$\Rightarrow 4t-4=0\Rightarrow t=1\text{sec}$
And, we have $x\left(t\right)=2t^2-4t+4$
So, at $t=0,x=4\text{m}$,
at $t=1,x=2\times 1^2-4\times 1+4=2\text{m}$
at $t=6,x=2\times 6^2-4\times 6+4=52\text{m}$
The motion of the particle can be represented as shown

So, the average speed of the particle is $\frac{2\times 2+48}{6}=\frac{26}{3}\text{m/s}$