An advertisement states that a 1,250kg car can go from 0 to 60 miles per hour in 3.75 seconds.
-Determine its acceleration
-Calculate net force required to give the car the acceleration
-Find the normal force exerted by the road on the car
- The coefficient of friction of the car tires is .80
Calculate the maximum force of friction between the car tires and the road
- Explain if the advertisement is possible
-Determine its acceleration
-Calculate net force required to give the car the acceleration
-Find the normal force exerted by the road on the car
- The coefficient of friction of the car tires is .80
Calculate the maximum force of friction between the car tires and the road
- Explain if the advertisement is possible
-
60 miles/h = 26.67 m/s
a = delta(v)/delta(t) = (26m/s) / (3.75 s) = 7.11 m/s^2
F = m a = 1250 kg * 7.11 m/s^2 = 8888.9 N
Normal force equals the weight of the car:
Fn = m g = 1250 kg * 9.81 m/s^2= 11263 N
Max friction equals
F_fricmax = mu * Fn = 0.80 * 11263 N = 9810 N
So the advertisement is possible, but only just. The maximal frictional force is just enough to deliver the force needed for accelerating at 7.11 m/s^2 .
(A quicker check would be that 7.11 m/s^2 is less than 0.80 g , because the mass drops out of the equation.)
a = delta(v)/delta(t) = (26m/s) / (3.75 s) = 7.11 m/s^2
F = m a = 1250 kg * 7.11 m/s^2 = 8888.9 N
Normal force equals the weight of the car:
Fn = m g = 1250 kg * 9.81 m/s^2= 11263 N
Max friction equals
F_fricmax = mu * Fn = 0.80 * 11263 N = 9810 N
So the advertisement is possible, but only just. The maximal frictional force is just enough to deliver the force needed for accelerating at 7.11 m/s^2 .
(A quicker check would be that 7.11 m/s^2 is less than 0.80 g , because the mass drops out of the equation.)
-
It's amazing!I find this.http://bufen.info/62941/cars-tyres