Question

If velocity of a particle is $v\left(t\right)=(15-3t)\text{m/s}$, the average speed of the particle in $6\text{sec}$ is

• A. $5\text{m/s}$
• B. $5.5\text{m/s}$
• C. $6.5\text{m/s}$
• D. $6\text{m/s}$

Answer 1

The correct option is C $6.5\text{m/s}$

Given, $v\left(t\right)=15-3t$, 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.
$\Rightarrow 15-3t=0\Rightarrow t=5\text{sec}$
Also, we know that
$x_f-x_i={\int }_0^{t}v\left(t\right)dt={\int }_0^t(15-3t)dt$
$\Rightarrow x_f-x_i=15t-\frac{3t^2}{2}$
Let particle starts from origin i.e $x_i=0$
So, at $t=0,x_f=0$,
at $t=5,x_f=15\times 5-\frac{3\times 5^2}{2}=\frac{75}{2}=37.5\text{m}$
at $t=6,x_f=15\times 6-\frac{3\times 6^2}{2}=36\text{m}$
The motion of the particle can be represented as shown

So, the average speed of the particle is $\frac{37.5+1.5}{6}=\frac{39}{6}=6.5\text{m/s}$