Question

If a particle has velocity $v\left(t\right)=t^2-7t+12$, the average speed in $6\text{sec}$ is

• A. $\frac{50}{6}\text{m/s}$
• B. $\frac{55}{18}\text{m/s}$
• C. $\frac{61}{18}\text{m/s}$
• D. $\frac{43}{6}\text{m/s}$

Answer 1

The correct option is B $\frac{55}{18}\text{m/s}$

Given, $v\left(t\right)=t^2-7t+12$, 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.
$\Rightarrow t^2-7t+12=0\Rightarrow (t-3)(t-4)=0\Rightarrow t=3,4\text{sec}$
Also, we know that
$x_f-x_i={\int }_0^{t}v\left(t\right)dt={\int }_0^t(t^2-7t+12)dt$
$\Rightarrow x_f-x_i=\frac{t^3}{3}-\frac{7t^2}{2}+12t$
Let particle starts from origin i.e $x_i=0$
So, at $t=0,x_f=0$,
at $t=3,x_f=\frac{3^3}{3}-\frac{7\times 3^2}{2}+12\times 3=\frac{27}{2}\text{m}$
at $t=4,x_f=\frac{4^3}{3}-\frac{7\times 4^2}{2}+12\times 4=\frac{40}{3}\text{m}$
at $t=6,x_f=\frac{6^3}{3}-\frac{7\times 6^2}{2}+12\times 6=18\text{m}$
The motion of the particle can be represented as shown

So, the average speed of the particle is $\frac{\frac{27}{2}+(\frac{27}{2}-\frac{40}{3})+(18-\frac{40}{3})}{6}=\frac{55}{18}\text{m/s}$