Question

Spot the dog is running after a ball.In 22m,he accelerates at a constant rate,and his velocity increases from 5.0m/s to 11m/s. What was his acceleration.

Answer 1

Here it is given that speed of the dog will increase from 5 m/s to 11 m/s in order to cover the distance of 22 m

so here we can use kinematics to find the acceleration

$v_f = 11 m/s$

$v_i = 5 m/s$

$d = 22 m$

now we will have

$v_f^2 - v_i^2 = 2 a d$

$11^2 - 5^2 = 2(a)(22)$

$a = \frac{11^2 - 5^2}{44}$

$a = 2.18 m/s^2$

so here it will accelerate with a = 2.18 m/s^2