Question

If velocity of a particle is $v\left(t\right)=(5t-10)\text{m/s}$, the average speed and magnitude of average velocity of the particle in $4\text{sec}$ respectively are

• A. $5\text{m/s},4\text{m/s}$
• B. $5\text{m/s},0\text{m/s}$
• C. $3\text{m/s},5\text{m/s}$
• D. $0\text{m/s},3\text{m/s}$

Answer 1

The correct option is B $5\text{m/s},0\text{m/s}$

Given, $v\left(t\right)=5t-10$, 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.
$\Rightarrow 5t-10=0\Rightarrow t=2\text{sec}$
Also, we know that
$x_f-x_i={\int }_0^{t}v\left(t\right)dt={\int }_0^t(5t-10)dt$
$\Rightarrow x_f-x_i=\frac{5t^2}{2}-10t$
Let particle starts from origin i.e $x_i=0$
So, at $t=0,x_f=0$,
at $t=2,x_f=\frac{5\times 2^2}{2}-10\times 2=-10\text{m}$
at $t=4,x_f=\frac{5\times 4^2}{2}-10\times 4=0\text{m}$
The motion of the particle can be represented as shown

So, the average speed of the particle is $\frac{2\times |-10|}{4}=\frac{20}{4}=5\text{m/s}$
and, average velocity is $\frac{0}{4}=0\text{m/s}$