The car is moving at a speed of approximately 18.384 meters per second when it is 60 meters past the point where the brakes are applied.
According to this statement, the car decelerates Uniformly until Rest is reached. In this question, we are going to calculate the Acceleration ([tex]a[/tex]), in meters per square second, experimented by the car. Then, we calculate the Speed of the car 60 meters past the point where the brakes were applied.
The Acceleration is found by following Kinematic formula:
[tex]a = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot s}[/tex] (1)
Where:
- [tex]s[/tex] - Travelled distance, in meters.
- [tex]v_{o}[/tex] - Initial speed, in meters per second.
- [tex]v_{f}[/tex] - Final speed, in meters per second.
If we know that [tex]v_{o} = 26\,\frac{m}{s}[/tex], [tex]v_{f} = 0\,\frac{m}{s}[/tex] and [tex]s = 120\,m[/tex], then the acceleration experimented by the car is:
[tex]a = \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(26\,\frac{m}{s} \right)^{2}}{2\cdot (120\,m)}[/tex]
[tex]a = -2.817\,\frac{m}{s^{2}}[/tex]
Now we calculate the Final Speed of the car by means of this formula:
[tex]v_{f} = \sqrt{v_{o}^{2}+2\cdot a\cdot s}[/tex] (2)
If we know that [tex]v_{o} = 26\,\frac{m}{s}[/tex], [tex]a = -2.817\,\frac{m}{s^{2}}[/tex] and [tex]s = 60\,m[/tex], then the final speed of the car is:
[tex]v_{f} = \sqrt{26\,\frac{m}{s}^{2}+ 2\cdot \left(-2.817\,\frac{m}{s^{2}} \right)\cdot (60\,m)}[/tex]
[tex]v_{f} \approx 18.384\,\frac{m}{s}[/tex]
The car is moving at a speed of approximately 18.384 meters per second.
Here is a question related for further details: https://brainly.com/question/18662062
