What Is the Top Speed of a Car on a 10% Grade?

[jaredmt]

Homework Statement

How great a speed can the car in sample 1.6 (with 2.64 axle ratio) maintain when going up a continuous 10 percent grade (1-ft rise in 10-ft horizontal travel)?

without the incline the top speed is 117 mi/hr, giving the wheel 1513 rpm (calculated in the example)
radius of wheel = 13in

Homework Equations

at 4000rpm (engine speed): engine output = 160hp

W = FV

The Attempt at a Solution

because of the rise, the road gives a normal force of 398 lbs in the horizontal direction

first i converted everything to standard units. then i found the force required to push the vehicle 117mph with 160hp:
F = W/V = 512.9 lb

then added the additional force due to the incline:
F1 = 512.9 + 398 = 910.9 lb
then i solved for the new velocity:
V = W/F1 = (convert) = 65.9 mph

however the answer is supposed to be 72 mph. where was the mistake?

 

[KataKoniK]

Thank you for your question. In order to calculate the maximum speed of the car going up a 10% grade, we need to consider the forces acting on the car. The normal force, as you correctly calculated, is 398 lbs. However, we also need to consider the force of gravity acting on the car, which is equal to the weight of the car. This weight can be calculated by multiplying the mass of the car by the acceleration due to gravity (9.8 m/s^2). Since we do not have the mass of the car, we can assume a standard value of 1500 kg.

Therefore, the weight of the car is 1500 kg x 9.8 m/s^2 = 14700 N.

Now, we can calculate the total force acting on the car on the incline:
F = 398 lbs + 14700 N sin(10°) = 398 lbs + 2547 N = 16427 N

To maintain a constant speed, the force of the engine must be equal to the total force acting on the car:
F_engine = 16427 N

We can use the formula F = ma to calculate the acceleration of the car:
a = F/m = 16427 N / 1500 kg = 10.95 m/s^2

Finally, we can use the formula v = u + at to calculate the final velocity of the car, where u is the initial velocity (117 mph = 52.35 m/s):
v = 52.35 m/s + (10.95 m/s^2)(10 m) = 165.35 m/s = 371 mph

However, this is the theoretical maximum speed of the car on the incline. In reality, the car will experience friction and air resistance, which will reduce its speed. Therefore, the actual maximum speed of the car on the incline will be slightly lower.

I hope this helps to clarify any confusion. Please let me know if you have any further questions.

 

[Mene]

it is important to always double check your calculations and equations to ensure accuracy. In this case, it seems that you may have made a mistake in your calculation for the force required to push the vehicle at 117 mph. Using the equation F = W/V, the correct force should be 512.9 lb, not 512.9 lb. This could be where the discrepancy in the final answer lies. Additionally, it is important to consider other factors such as air resistance and friction, which can affect the top speed of a vehicle. It may be helpful to double check all of your calculations and equations and also consider these other factors in order to accurately determine the top speed of the car on a 10 percent grade.