Question

1. saagar gives a certain teacher some sass followed by an eye - roll. suddenly, a chankla is thrown at him. the flip - flop moves from rest to 12km/hr when it strikes saagar in the back of the head. the flight of the chankla takes only 1.2 seconds. what is the average acceleration of the chankla?

Explanation:

Step1: Convert speed to m/s

The final speed $v = 12\ km/h=\frac{12\times1000}{3600}\ m/s=\frac{10}{3}\ m/s$, initial speed $u = 0\ m/s$, and time $t=1.2\ s$.

Step2: Use acceleration formula

The formula for average acceleration is $a=\frac{v - u}{t}$. Substitute $u = 0\ m/s$, $v=\frac{10}{3}\ m/s$ and $t = 1.2\ s$ into the formula. So $a=\frac{\frac{10}{3}-0}{1.2}=\frac{10}{3\times1.2}=\frac{10}{3.6}=\frac{100}{36}=\frac{25}{9}\approx2.78\ m/s^{2}$.

Answer:

$\frac{25}{9}\ m/s^{2}\approx2.78\ m/s^{2}$