Answer :
To determine Kyle's kinetic energy while jogging, we can use the kinetic energy formula:
[tex]\[ KE = \frac{1}{2}mv^2 \][/tex]
where:
- [tex]\( KE \)[/tex] is the kinetic energy,
- [tex]\( m \)[/tex] is the mass (in kilograms),
- [tex]\( v \)[/tex] is the velocity (in meters per second).
Given:
- The mass [tex]\( m \)[/tex] is [tex]\( 54 \, \text{kg} \)[/tex],
- The velocity [tex]\( v \)[/tex] is [tex]\( 3 \, \text{m/s} \)[/tex].
Substitute the given values into the formula:
[tex]\[ KE = \frac{1}{2} \times 54 \, \text{kg} \times (3 \, \text{m/s})^2 \][/tex]
First, calculate the square of the velocity:
[tex]\[ (3 \, \text{m/s})^2 = 9 \, \text{m}^2/\text{s}^2 \][/tex]
Next, multiply the mass by this squared velocity:
[tex]\[ 54 \, \text{kg} \times 9 \, \text{m}^2/\text{s}^2 = 486 \, \text{kg} \cdot \text{m}^2/\text{s}^2 \][/tex]
Since the formula includes a factor of [tex]\(\frac{1}{2}\)[/tex], we need to divide this product by 2:
[tex]\[ KE = \frac{1}{2} \times 486 \, \text{kg} \cdot \text{m}^2/\text{s}^2 = 243 \, \text{kg} \cdot \text{m}^2/\text{s}^2 \][/tex]
Therefore, Kyle's kinetic energy is:
[tex]\[ 243 \, \text{J} \][/tex]
So, the correct answer is:
[tex]\[ 243 \, \text{J} \][/tex]
[tex]\[ KE = \frac{1}{2}mv^2 \][/tex]
where:
- [tex]\( KE \)[/tex] is the kinetic energy,
- [tex]\( m \)[/tex] is the mass (in kilograms),
- [tex]\( v \)[/tex] is the velocity (in meters per second).
Given:
- The mass [tex]\( m \)[/tex] is [tex]\( 54 \, \text{kg} \)[/tex],
- The velocity [tex]\( v \)[/tex] is [tex]\( 3 \, \text{m/s} \)[/tex].
Substitute the given values into the formula:
[tex]\[ KE = \frac{1}{2} \times 54 \, \text{kg} \times (3 \, \text{m/s})^2 \][/tex]
First, calculate the square of the velocity:
[tex]\[ (3 \, \text{m/s})^2 = 9 \, \text{m}^2/\text{s}^2 \][/tex]
Next, multiply the mass by this squared velocity:
[tex]\[ 54 \, \text{kg} \times 9 \, \text{m}^2/\text{s}^2 = 486 \, \text{kg} \cdot \text{m}^2/\text{s}^2 \][/tex]
Since the formula includes a factor of [tex]\(\frac{1}{2}\)[/tex], we need to divide this product by 2:
[tex]\[ KE = \frac{1}{2} \times 486 \, \text{kg} \cdot \text{m}^2/\text{s}^2 = 243 \, \text{kg} \cdot \text{m}^2/\text{s}^2 \][/tex]
Therefore, Kyle's kinetic energy is:
[tex]\[ 243 \, \text{J} \][/tex]
So, the correct answer is:
[tex]\[ 243 \, \text{J} \][/tex]
