Question

Complete the following table:

Car Model	Driver X Reaction time 0.20 s	Driver Y Reaction time 0.30 s
A (deceleration on hard braking = 6.0 m/s2)	Speed = 54 km/h Braking distance a = ............ Total stopping distance b = ............	Speed = 72 km/h Braking distance c = ........... Total stopping distance d = ............
B (deceleration on hard braking = 7.5 m/s2)	Speed = 54 km/h Breaking distance e = ........... Total stopping distance f = ............	Speed 72 km/h Braking distance g = ............. Total stopping distance h = ............

Answer 1

Braking distance: Distance travelled after the brakes are applied.
Total stopping distance = Braking distance + Distance travelled in the reaction time

Case A:
Deceleration = 6.0 m/s²
For driver X:
Initial velocity, u = 54 km/h = 15 m/s
Final velocity, v = 0
Braking distance, $a=\frac{0^2-{15}^2}{2\left(-6\right)}\approx 19\mathrm{m}$
Distance travelled in the reaction time = 15 × 0.20 = 3 m
Total stopping distance, b = 19 + 3 = 22 m

For driver Y:
Initial velocity, u = 72 km/h = 20 m/s
Final velocity, v = 0
Braking distance, $c=\frac{0^2-{20}^2}{2\left(-6\right)}\approx 33\mathrm{m}$
Distance travelled in the reaction time = 20 × 0.30 = 6 m
Total stopping distance, d = 33 + 6 = 39 m

Case B:
Deceleration = 6.0 m/s²
Now, we have:
e = 15 m
f = 18 m
g = 27 m
h = 33 m

Car Model	Driver X Reaction Time = 0.20 s	Driver Y Reaction Time = 0.30 s
A (deceleration on hard braking = 6.0 m/s2)	Speed = 54 km/h Braking distance, a = 19 m Total stopping distance, b = 22 m	Speed = 72 km/h Braking distance, c = 33 m Total stopping distance, d = 39 m
B (deceleration on hard braking = 7.5 m/s2)	Speed = 54 km/h Breaking distance, e = 15 m Total stopping distance, f = 18 m	Speed = 72 km/h Braking distance, g = 27 m Total stopping distance, h = 33 m