The distance,s metres in which a car can stop is related to its speed, s=pu+qv² where p and q are constants. A car travelling at 10km/h can stop in a distance 5 meters and at 20km/h in a distance 12 meters. Calculate the distance in which the car travelling at 30km/h can escape.​



Answer :

Answer:

20.97 meters to stop.

Step-by-step explanation:

Given:

  • Stopping Distance (s) = pu + qv²

Where:

  • s is the stopping distance in meters
  • v is the speed in km/h
  • p and q are constants we need to find

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.