The motion of the car has a constant acceleration. In this case we will use the expression of the acceleration and the expression of distance for the uniformly varied movement.
The acceleration is defined as:
a = (v – v0)/t
v – v0 → Variation of speed, at time t
The speed variation, is from zero to 21 km/h in a time of 5.6 seconds.
21 km/h = (21)(1000/3600) = 5.83 m/s
Then the acceleration is:
a = (5.83 – 0)/5.6 = 1.04 m/s^2
Now, we apply the expression of distance:
d = v0t + at^2/2
d = 0 + (1.04)(5.6)^2/2
d = 16.3 m
Then, this car travels 16.3 m in a time of 5.6 s