To determine the time it takes for the baseball to reach its final velocity, we can use one of the fundamental kinematic equations that relates velocity, acceleration, and time.
The kinematic equation we use is:
[tex]\[ v = u + at \][/tex]
Where:
- [tex]\( v \)[/tex] is the final velocity
- [tex]\( u \)[/tex] is the initial velocity
- [tex]\( a \)[/tex] is the acceleration
- [tex]\( t \)[/tex] is the time
First, we identify the given values:
- The initial velocity ([tex]\( u \)[/tex]) is [tex]\( -7.00 \, \text{m/s} \)[/tex]
- The final velocity ([tex]\( v \)[/tex]) is [tex]\( -12.5 \, \text{m/s} \)[/tex]
- The acceleration ([tex]\( a \)[/tex]) is [tex]\( 9.8 \, \text{m/s}^2 \)[/tex]
We need to solve for the time ([tex]\( t \)[/tex]).
Rearranging the kinematic equation to solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{v - u}{a} \][/tex]
Substituting the given values into the equation:
[tex]\[ t = \frac{-12.5 \, \text{m/s} - (-7.00 \, \text{m/s})}{9.8 \, \text{m/s}^2} \][/tex]
Performing the subtraction in the numerator:
[tex]\[ t = \frac{-12.5 \, \text{m/s} + 7.00 \, \text{m/s}}{9.8 \, \text{m/s}^2} \][/tex]
[tex]\[ t = \frac{-5.5 \, \text{m/s}}{9.8 \, \text{m/s}^2} \][/tex]
Now, performing the division:
[tex]\[ t \approx -0.561 \, \text{s} \][/tex]
So, the time it takes for the baseball to reach its final velocity is approximately [tex]\( t = -0.561 \, \text{s} \)[/tex].
This negative value indicates that the final velocity was reached before the time considered as a reference point (if we are considering time [tex]\( t = 0 \)[/tex] as some initial moment).
