Question

If velocity of a particle is $v\left(t\right)=2t-4\text{m/s}$, the average speed and magnitude of average velocity of the particle in $10\text{sec}$ respectively are

• A. $4.8\text{m/s},4\text{m/s}$
• B. $6\text{m/s},6\text{m/s}$
• C. $6.8\text{m/s},6\text{m/s}$
• D. $5.2\text{m/s},4.4\text{m/s}$

Answer 1

The correct option is C $6.8\text{m/s},6\text{m/s}$

Given, $v\left(t\right)=2t-4$, 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.
$\Rightarrow 2t-4=0\Rightarrow t=2\text{sec}$
Also, we know that
$x_f-x_i={\int }_0^{t}v\left(t\right)dt={\int }_0^t(2t-4)dt$
$\Rightarrow x_f-x_i=t^2-4t$
Let particle starts from origin i.e $x_i=0$
So, at $t=0,x_f=0$,
at $t=2,x_f=2^2-4\times 2=-4\text{m}$
at $t=10,x_f={10}^2-4\times 10=60\text{m}$
The motion of the particle can be represented as shown

So, the average speed of the particle is $\frac{|-4|+|-4|+60}{10}=\frac{68}{10}=6.8\text{m/s}$
and, average velocity is $\frac{60}{10}=6\text{m/s}$