Question

If a particle has velocity $v\left(t\right)=t^2-5t+6$, the average velocity in $5\text{sec}$ is

• A. $\frac{13}{14}\text{m/s}$
• B. $\frac{14}{15}\text{m/s}$
• C. $\frac{11}{12}\text{m/s}$
• D. $\frac{7}{9}\text{m/s}$

Answer 1

The correct option is C $\frac{11}{12}\text{m/s}$

Given, $v\left(t\right)=t^2-5t+6$,
Also, we know that
$x_f-x_i={\int }_0^{t}v\left(t\right)dt={\int }_0^t(t^2-5t+6)dt$
$\Rightarrow x_f-x_i=\frac{t^3}{3}-\frac{5t^2}{2}+6t$
for average velocity, we need to find the final position of the particle at time $t=5\text{sec}$
Let particle starts from origin i.e $x_i=0$
So, at $t=0,x_f=0$,
at $t=5,x_f=\frac{5^3}{3}-5\times \frac{5^2}{2}+6\times 5=\frac{55}{6}\text{m}$
Hence, average velocity is $\frac{\frac{55}{6}}{10}=\frac{11}{12}\text{m/s}$