Question

A ball rolled 12.0 m [E] in 10.0 s, hit an obstacle, and rolled straight back. After the collision, the ball rolled 8.00 m [W] in 6.00 s. What was the average velocity of the ball?

Answer 1

Answer:

the average velocity of the ball is 5 m/s.

Explanation:

Given;

initial position of the ball, x₁ = 12 m East

final position of the ball, x₂ = 8 m West

initial time of motion, t₁ = 10.0 s

final time of motion, t₂ = 6.0 s

The average velocity is calculated as follows;

$v = \frac{\Delta x}{\Delta t} = \frac{x_1 \ - \ x_2}{t_1 \ - \ t_2}$

let the Eastward direction be positive

Let the westward direction be negative

$\frac{x_1 \ - \ x_2}{t_1 \ - \ t_2} = \frac{12 \ - \ (-8)}{10 -6}= \frac{20}{4} = 5 \ m/s$

Therefore, the average velocity of the ball is 5 m/s.