Question

If velocity of a particle is $v\left(t\right)=(9-3t)\text{m/s}$, the average speed and magnitude of average velocity of the particle in $4\text{sec}$ respectively are

• A. $5.5\text{m/s},5\text{m/s}$
• B. $3.75\text{m/s},3\text{m/s}$
• C. $3.5\text{m/s},3\text{m/s}$
• D. $5.75\text{m/s},5\text{m/s}$

Answer 1

The correct option is B $3.75\text{m/s},3\text{m/s}$

Given, $v\left(t\right)=9-3t$, 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.
$\Rightarrow 9-3t=0\Rightarrow t=3\text{sec}$
Also, we know that
$x_f-x_i={\int }_0^{t}v\left(t\right)dt={\int }_0^t(9-3t)dt$
$\Rightarrow x_f-x_i=9t-\frac{3t^2}{2}$
Let particle starts from origin i.e $x_i=0$
So, at $t=0,x_f=0$,
at $t=3,x_f=9\times 3-\frac{3\times 3^2}{2}=\frac{27}{2}=13.5\text{m}$
at $t=4,x_f=9\times 4-\frac{3\times 4^2}{2}=12\text{m}$
The motion of the particle can be represented as shown

So, the average speed of the particle is $\frac{13.5+1.5}{4}=\frac{15}{4}=3.75\text{m/s}$
and, average velocity is $\frac{12}{4}=3\text{m/s}$