Question

A particle moves from $A$ to $B$ such that $x=t^2+t-3$. Its average velocity from $t=2s$ to $t=5s$ is

• A. $6m{s}^{-1}$
• B. $8m{s}^{-1}$
• C. $8.5m{s}^{-1}$
• D. $7m{s}^{-1}$

Answer 1

Answer: (B)

$8m{s}^{-1}$

Here, $x=t^2+t-3,t_2=5\&t_1=2$
so, $v_{av}=\frac{x\left(t_2\right)-x\left(t_1\right)}{{t_2-t}_1}=\frac{(5^2+5-3)-(2^2+2-3)}{5-2}=\frac{24}{3}=8m/s$