Question

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

• A. $\frac{55}{18}\text{m/s},3.5\text{m/s}$
• B. $\frac{71}{18}\text{m/s},4\text{m/s}$
• C. $\frac{51}{18}\text{m/s},5\text{m/s}$
• D. $\frac{55}{18}\text{m/s},3\text{m/s}$

Answer 1

The correct option is D $\frac{55}{18}\text{m/s},3\text{m/s}$

Given, $v\left(t\right)=t^2-5t+6$, 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.
$\Rightarrow t^2-5t+6=0\Rightarrow (t-2)(t-3)=0\Rightarrow t=2,3\text{sec}$
Also, we know that
$x_f-x_i={\int }_0^{t}v\left(t\right)dt={\int }_0^t(t^2-5t+6)dt$
$\Rightarrow x_f-x_i=\frac{t^3}{3}-\frac{5t^2}{2}+6t$
Let particle starts from origin i.e $x_i=0$
So, at $t=0,x_f=0$,
at $t=2,x_f=\frac{2^3}{3}-\frac{5\times 2^2}{2}+6\times 2=\frac{14}{3}\text{m}$
at $t=3,x_f=\frac{3^3}{3}-\frac{5\times 3^2}{2}+6\times 3=\frac{9}{2}\text{m}$
at $t=6,x_f=\frac{6^3}{3}-\frac{5\times 6^2}{2}+6\times 6=18\text{m}$
The motion of the particle can be represented as shown

So, the average speed of the particle is $\frac{\frac{14}{3}+(\frac{14}{3}-\frac{9}{2})+(18-\frac{9}{2})}{6}=\frac{55}{18}\text{m/s}$
and, average velocity is $\frac{18}{6}=3\text{m/s}$