Question

If velocity of a particle is $v\left(t\right)=8-2t\text{m/s}$, the average speed and magnitude of average velocity of the particle in $5\text{sec}$ respectively are

• A. $3\text{m/s},\frac{17}{5}\text{m/s}$
• B. $3\text{m/s},\frac{31}{5}\text{m/s}$
• C. $\frac{17}{5}\text{m/s},3\text{m/s}$
• D. $\frac{17}{5}\text{m/s},\frac{17}{5}\text{m/s}$

Answer 1

The correct option is C $\frac{17}{5}\text{m/s},3\text{m/s}$

Given, $v\left(t\right)=8-2t$, 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.
$\Rightarrow 8-2t=0\Rightarrow t=4\text{sec}$
Also, we know that
$x_f-x_i={\int }_0^{t}v\left(t\right)dt={\int }_0^t(8-2t)dt$
$\Rightarrow x_f-x_i=8t-t^2$
Let particle starts from origin i.e $x_i=0$
So, at $t=0,x_f=0$,
at $t=4,x_f=8\times 4-4^2=16\text{m}$
at $t=5,x_f=8\times 5-5^2=15\text{m}$
The motion of the particle can be represented as shown

So, the average speed of the particle is $\frac{16+1}{5}=\frac{17}{5}\text{m/s}$
and, average velocity is $\frac{15}{5}=3\text{m/s}$