Question

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

• A. $2.5\text{m/s}$
• B. $3.5\text{m/s}$
• C. $4.5\text{m/s}$
• D. $3\text{m/s}$

Answer 1

The correct option is A $2.5\text{m/s}$

As we know that the velocity of the particle as a function of time is given by
$\frac{dx}{dt}=\frac{d(t^2-6t)}{dt}$
$\Rightarrow v=2t-6$
Now, $v\left(t\right)=2t-6$, 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.
$\Rightarrow 2t-6=0\Rightarrow t=3\text{sec}$
And, we have $x\left(t\right)=t^2-6t$
So, at $t=0,x=0\text{m}$,
at $t=3,x=3^2-6\times 3=-9\text{m}$
at $t=4,x=4^2-6\times 4=-8\text{m}$
The motion of the particle can be represented as shown

So, the average speed of the particle is $\frac{|-9|+1}{4}=2.5\text{m/s}$