Question

Velocity vs time graph of a particle is shown in figure. If $v_{max}$ of particle is $a$ and $v_{avg}$ of particle is $b$, then

• A. $a=b$
• B. $a=2b$
• C. $a/b$ depends upon $t_1$ and $t_2$
• D. $b=2a$

Answer 1

The correct option is B $a=2b$

Displacement = area $=\frac{1}{2}\text{(height)(base)}$
$V_{avg}=\frac{\text{Displacement}}{\text{time}}=\frac{\frac{1}{2}\left(a\right)(t_1+t_2)}{(t_1+t_2)}=\frac{a}{2}$
$V_{max}=a$
$\begin{array}{c}{V}_{max}=a\\ V_{avg}=b=\frac{a}{2}\end{array}\}$
$a=2b$