Question

A particle moves along a straight line and its velocity depends on time as $v=4t-t^2.$Then for first $5\text{sec}$

• A. average velocity is $\frac{25}{3}\text{m/s}$
• B. average velocity is $10\text{m/s}$
• C. average velocity is $\frac{5}{3}\text{m/s}$
• D. average velocity is $\frac{13}{5}\text{m/s}$

Answer 1

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

Given, $v\left(t\right)=4t-t^2$, 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.
$\Rightarrow 4t-t^2=0\Rightarrow t=0,4\text{sec}$
Also, we know that
$x_f-x_i={\int }_0^{t}v\left(t\right)dt={\int }_0^t(4t-t^2)dt$
$\Rightarrow x_f-x_i=2t^2-\frac{t^3}{3}$
Let particle starts from origin i.e $x_i=0$
So, at $t=0,x_f=0$,
at $t=4,x_f=2\times 4^2-\frac{4^3}{3}=\frac{96-64}{3}=\frac{32}{3}\text{m}$
at $t=5,x_f=2\times 5^2-\frac{5^3}{3}=\frac{150-125}{3}=\frac{25}{3}\text{m}$
Hence, average velocity is $\frac{25}{3\times 5}=\frac{5}{3}\text{m/s}$