Question

An athlete takes $2s$ to reach his maximum speed $18{}^{km}/{}_{hr}$ after starting from rest. What is the magnitude of his average acceleration?

Answer 1

Hint Formula for average acceleration is
Average acceleration $=\dfrac{{{v}_{f}}+{{v}_{i}}}{2}$
Where ${{v}_{f}}=$ final velocity, and ${{v}_{i}}=$ initial velocity

Complete step-by-step solution :
Acceleration is a vector quantity that is defined as the rate at which an object changes its velocity.
Average acceleration is the final velocity per time taken.
$a=\dfrac{{{v}_{f}}-{{v}_{i}}}{t}$
Average acceleration$=\dfrac{{{v}_{f}}+{{v}_{i}}}{2}$
Initial velocity $\left( {{v}_{i}} \right)=0$
Final velocity $v_f= 18kmph$
$\begin{aligned} & {{v}_{f}}=18\times \dfrac{1000}{60\times 60}{m}/{\sec }\; \\\ & v_f=5m/sec \\\ \end{aligned}$
$t=0$ to$t=2$
Average acceleration$=\dfrac{{{v}_{f}}+{{v}_{i}}}{2}$
$\begin{aligned} & =\dfrac{18+0}{2} \\\ & 9kmph\\\ \end{aligned}$
Average acceleration $=\dfrac{{{v}_{f}}+{{v}_{i}}}{2}$
$\begin{aligned} & =\dfrac{5+0}{2} \\\ & =5{m}/{\sec }\; \\\ \end{aligned}$

Note: Student thought that the acceleration and average acceleration is same but for acceleration $\left( \dfrac{{{v}_{f}}-{{v}_{i}}}{t} \right)$ and average acceleration is $\left( \dfrac{{{v}_{f}}+{{v}_{i}}}{2} \right)$ . So carefully use the formula of both.