Question

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

a. what is the acceleration?
1m/s²
b. what is the final velocity after the 10 s?

Explanation:

Part a:

Step1: Recall the kinematic equation

The kinematic equation for displacement with initial velocity \( u = 0 \) (starts from rest) is \( s=ut+\frac{1}{2}at^{2} \). Here, \( s = 50\space m \), \( u = 0\space m/s \), \( t = 10\space s \), and we need to find \( a \).

Step2: Substitute values into the equation

Substitute \( u = 0 \), \( s = 50 \), \( t = 10 \) 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: \( 50=\frac{1}{2}\times a\times100 = 50a \).

Step3: Solve for \( a \)

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

Part b:

Step1: Recall the kinematic equation

The kinematic equation for final velocity \( v \) with initial velocity \( u = 0 \) is \( v=u + at \). We know that \( u = 0\space m/s \), \( a = 1\space m/s^{2} \), and \( t = 10\space s \).

Step2: Substitute values into the equation

Substitute \( u = 0 \), \( a = 1 \), and \( t = 10 \) into \( v=u + at \). We get \( v=0+1\times10=10\space m/s \).

Answer:

(Part a):
The acceleration is \( \boldsymbol{1\space m/s^{2}} \)