The keyword slows down “uniformly” means that the car had
a constant deceleration. Therefore we can use the formulas for linear motion.
a = (vf – vi) / t --->
1
s = vi t + 0.5 a t^2 --->
2
where a is acceleration, v is velocity with notation f and
i for final and initial, s is the distance travelled, and t stands for time
First we solve for acceleration using equation 1:
a = (0 – 21) / 6
a = - 3.5 m / s^2 (negative
means deceleration)
Then we can now calculate for the value of s using
equation 2:
s = 21 (6) + 0.5 (-3.5) (6)^2
s = 63 m
Therefore the car travelled 63 m before it came to a stop.
