Answer: 12.24m/s
According to kinematics this situation is described as a uniformly accelerated rectilinear motion. This means the acceleration while the car is in motion is constant.
Now, among the equations related to this type of motion we have the following that relates the velocity with the acceleration and the distance traveled:
[tex]V_{f}^{2}-V_{o}^{2}=2.a.d[/tex] (1)
Where:
[tex]V_{f}[/tex] is the Final Velocity of the car. We are told "the car comes to a stop after travelling", this means it is 0.
[tex]V_{o}[/tex] is the Initial Velocity, the value we want to find
[tex]a[/tex] is the constant acceleration of the car (the negative sign means the car is decelerating)
[tex]d[/tex] is the distance traveled by the car
Now, let's substitute the known values in equation (1) and find [tex]V_{o}[/tex]:
[tex]0-V_{o}^{2}=2(-5m/{s}^{2})(15m)[/tex] (2)
[tex]-V_{o}^{2}=-150{m}^{2}/{s}^{2}[/tex] (3)
Multiplying by -1 on both sides of the equation:
[tex]V_{o}^{2}=150{m}^{2}/{s}^{2}[/tex] (4)
[tex]V_{o}=\sqrt{150{m}^{2}/{s}^{2}}[/tex] (5)
Finally:
[tex]V_{o}=12.24m/s[/tex] >>>This is the Initial velocity of the car
