Answer:
x = 45 [m]
Explanation:
To solve this problem we must use the following kinematics equation:
[tex]v_{f}^{2} = v_{i}^{2}-(2*a*x)[/tex]
where:
Vf = final velocity = 0
Vi = initial velocity = 15 [m/s]
a = desacceleration = 2.5 [m/s^2]
x = distance [m]
Note: the negative sign in the equation above, is because the car slows down.
0^2 = 15^2 - (2*2.5*x)
2*2.5*x = 15^2
x = 45 [m]
