Answer:
20.97 meters to stop.
Step-by-step explanation:
Given:
Where:
From the first equation since p and q are constants we can derive that s is partly constant and varies with v^2
Converting the unit of speed to the unit of distance measured for simplicity
10 km/h ≈ 2.78 m/s
20 km/h ≈ 5.56 m/s
30 km/h ≈ 8.33 m/s
s = 5, v = u = 2.78 and s = 12, v = u = 5.56
5 = 2.78p + q(2.78)² --(1)
12 = 5.56p + q(5.56)² --(2)
From equation 1 make p the subject of the formula
5 = 2.78p
5 - 7.7284q = 2.78p
p = (5 - 7.7284q) / 2.78 --(3)
Substitute equation 3 for equation 2
12 = 5.56(5 - 7.7284q) / 2.78 + q(5.56)²
12 = 5.56(5 - 7.7284q) / 2.78 + 30.9136q
12 = (27.8 - 42.9699q) / 2.78 + 30.9136q
12 = 10 - 15.4567q + 30.9136q
30.9136q - 15.4567q = 12 - 10
15.4569q = 2
q = 2/15.4569
q = 0.1293
Find p by substituting q back into equation (3):
p = (5 - 7.7284 * 0.1293) / 2.78
p = (5 - 0.9993) / 2.78
p = 4.0007 / 2.78
p ≈ 1.4391
Now we have both constants: p ≈ 1.4391 and q ≈ 0.1293
To find the stopping distance at 30 km/h (8.33 m/s):
s = pu + qv²
s = 1.4391 * 8.33 + 0.1293 * (8.33)²
s = 11.9877 + 8.9795
s ≈ 20.97 meters
Therefore, a car traveling at 30 km/h would require approximately 20.97 meters to stop.