Stopping Distance: Reaction Time, Wet Roads and a Misleading Claim
A 1250 kg car is travelling at 22 m s⁻¹ on a dry, level road. The driver's reaction time, measured in a separate test, is 0.75 s. The brakes can produce a constant deceleration; a test from 22 m s⁻¹ gives a braking distance of 33 m. The driver sees a child step into the road 45 m ahead and claims, 'I can stop in time because my braking distance is only 33 m, which is less than 45 m.' The road is straight and the car does not skid. On a wet day the coefficient of friction between the tyres and the road is 0.60 times its dry value. The speedometer has an uncertainty of ±1 m s⁻¹, and the reaction time could be as much as 0.90 s for an alert but inexperienced driver. The car's centre of mass is low and the road is horizontal.
1. Calculate the dry-road deceleration and the total stopping distance from the moment the child is seen. Decide whether the driver's claim is correct, and estimate the speed at which the car would reach the child if it continued braking.
Read the worked answer
Using (v^2=u^2+2as) for the braking phase with (v=0), (u=22\ \text{m s}^{-1}) and (s=33\ \text{m}):
[ a=\frac{0-22^2}{2(33)}=-7.33\ \text{m s}^{-2} ]
So the deceleration is about (7.3\ \text{m s}^{-2}).
Reaction distance:
[ s_{\text{react}}=ut=(22)(0.75)=16.5\ \text{m} ]
Total stopping distance:
[ s_{\text{total}}=16.5+33=49.5\ \text{m} ]
This is greater than 45 m, so the claim is wrong. The driver has confused braking distance with total stopping distance.
At the child's position, only (45-16.5=28.5\ \text{m}) is available for braking. Then
[ v^2=22^2+2(-7.33)(28.5)=484-417.8=66.2 ]
[ v\approx8.1\ \text{m s}^{-1} ]
The car would still be moving at about 8 m s⁻¹ (about 29 km h⁻¹) when it reached the child.
2. On the wet road the friction coefficient is 0.60 times the dry value. Estimate the maximum initial speed at which the car could just stop in 45 m if the reaction time is 0.90 s. Comment on the assumptions in your calculation.
Read the worked answer
Dry friction coefficient:
[ \mu_{\text{dry}}=\frac{7.33}{9.81}=0.747 ]
Wet value:
[ \mu_{\text{wet}}=0.60(0.747)=0.448 ]
Wet deceleration:
[ a_{\text{wet}}=\mu_{\text{wet}}g=0.448(9.81)=4.40\ \text{m s}^{-2} ]
Let the maximum speed be (u). Reaction distance is (0.90u). Braking distance is (u^2/(2a)):
[ 0.90u+\frac{u^2}{2(4.40)}=45 ]
[ \frac{u^2}{8.80}+0.90u-45=0 ]
[ u^2+7.92u-396=0 ]
[ u=\frac{-7.92+\sqrt{7.92^2+4(396)}}{2}\approx16.3\ \text{m s}^{-1} ]
So the maximum speed is about 16 m s⁻¹, well below 22 m s⁻¹.
The calculation assumes constant deceleration, a fixed reaction time, and that the tyres operate at the limit of friction without skidding. In practice, weight transfer, road texture, water depth and driver alertness vary, so a real safety margin is needed.
3. A student says that because the wet friction coefficient is 0.60 of the dry value, the maximum safe speed should be 0.60 × 22 = 13.2 m s⁻¹. Evaluate this reasoning using the braking-distance equation.
Read the worked answer
The student has assumed that speed scales in direct proportion to the friction coefficient. That is incorrect.
For the same braking distance (s), (u^2=2\mu gs), so
[ u\propto\sqrt{\mu} ]
If (\mu) falls to 0.60 of its dry value, the speed that can be shed over the same braking distance falls to
[ \sqrt{0.60}=0.775 ]
of its dry value:
[ u_{\text{wet}}\approx0.775(22)=17.0\ \text{m s}^{-1} ]
Allowing a longer reaction time of 0.90 s reduces this further to about 16.3 m s⁻¹, as calculated above.
The student's value of 13.2 m s⁻¹ is therefore over-conservative as a friction-scaling argument, although a real speed limit would also include reaction-time and safety-margin considerations.