Question

The position of the particle is given by $x=t^2-4t+6$ in $\text{m}$. The average speed of the particle in $5\text{sec}$ is

• A. $\frac{9}{5}\text{m/s}$
• B. $\frac{13}{5}\text{m/s}$
• C. $\frac{11}{5}\text{m/s}$
• D. $\frac{17}{5}\text{m/s}$

Answer 1

The correct option is B $\frac{13}{5}\text{m/s}$

As we know that the velocity of the particle as a function of time is given by
$\frac{dx}{dt}=\frac{d(t^2-4t+6)}{dt}$
$\Rightarrow v=2t-4$
Now, $v\left(t\right)=2t-4$, 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.
$\Rightarrow 2t-4=0\Rightarrow t=2\text{sec}$
And, we have $x\left(t\right)=t^2-4t+6$
So, at $t=0,x=6\text{m}$,
at $t=2,x=2^2-4\times 2+6=2\text{m}$
at $t=5,x=5^2-4\times 5+6=11\text{m}$
The motion of the particle can be represented as shown

So, the average speed of the particle is $\frac{2\times 4+5}{5}=\frac{13}{5}\text{m/s}$