Question

1. a baseball starts from rest and rolls 50 m down a hill in 10 s.

a. what is the acceleration?

Explanation:

Step1: Recall the kinematic equation

The kinematic equation for displacement with constant acceleration starting from rest (\(u = 0\)) is \(s=ut+\frac{1}{2}at^{2}\). Here, \(s = 50\space m\), \(u = 0\space m/s\) (starts from rest), and \(t = 10\space s\).

Step2: Substitute the values into the equation

Substitute \(u = 0\), \(s = 50\), and \(t = 10\) into \(s=ut+\frac{1}{2}at^{2}\). We get \(50=0\times10+\frac{1}{2}\times a\times(10)^{2}\). Simplify the right - hand side: \(50 = 0+\frac{1}{2}\times a\times100\), which is \(50 = 50a\).

Step3: Solve for acceleration \(a\)

Divide both sides of the equation \(50 = 50a\) by 50. We have \(a=\frac{50}{50}=1\space m/s^{2}\).

Answer:

The acceleration of the baseball is \(1\space m/s^{2}\)