Question

If a particle has velocity $v\left(t\right)=t^2-6t+8$, the average speed in $5\text{sec}$ is

• A. $\frac{8}{5}\text{m/s}$
• B. $\frac{23}{15}\text{m/s}$
• C. $\frac{28}{15}\text{m/s}$
• D. $\frac{19}{5}\text{m/s}$

Answer 1

The correct option is C $\frac{28}{15}\text{m/s}$

Given, $v\left(t\right)=t^2-6t+8$, 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.
$\Rightarrow t^2-6t+8=0\Rightarrow (t-2)(t-4)=0\Rightarrow t=2,4\text{sec}$
Also, we know that
$x_f-x_i={\int }_0^{t}v\left(t\right)dt={\int }_0^t(t^2-6t+8)dt$
$\Rightarrow x_f-x_i=\frac{t^3}{3}-3t^2+8t$
Let particle starts from origin i.e $x_i=0$
So, at $t=0,x_f=0$,
at $t=2,x_f=\frac{2^3}{3}-3\times 2^2+8\times 2=\frac{20}{3}\text{m}$
at $t=4,x_f=\frac{4^3}{3}-3\times 4^2+8\times 4=\frac{16}{3}\text{m}$
at $t=5,x_f=\frac{5^3}{3}-3\times 5^2+8\times 5=\frac{20}{3}\text{m}$
The motion of the particle can be represented as shown

So, the average speed of the particle is $\frac{\frac{20}{3}+2\times (\frac{20}{3}-\frac{16}{3})}{5}=\frac{28}{15}\text{m/s}$