Calculating Car Acceleration on a 3% Slope at 88.5 km/h

[Apprentice123]

A car is rising an excerpt in slope of 2%, at a constant speed of 88,5 km/h.
If the driver does not change the motion, or changing the pressure of your foot on the accelerator, which should be the acceleration of the car down the stretch of 3%?Answer
0,491 m/s^2

 

[kNYsJakE]

A car is rising an excerpt in slope of 2%, at a constant speed of 80,5 km/h.
If the driver does not change the motion, or changing the pressure of your foot on the accelerator, which should be the acceleration of the car down the stretch of 3%?


Answer
0,491 m/s^2

People can't help you unless you post your attempted solutions. It's a place where people help you on homework, not do your homework for you.

 

[Apprentice123]

People can't help you unless you post your attempted solutions. It's a place where people help you on homework, not do your homework for you.

I'm not asking you to resolve. I would like an explanation of the exercise

 

[Apprentice123]

It is:

I have a triangle of sides 2 and 100. Find the angle with the floor.

theta = 1,145

velocity x sin(theta) = 24,58 x sin(1,145) = 0,4911 m/s^2.

correct ? I did not use the slope of 3%

 

[cepheid]

Your value for the angle of the incline (on the upward slope) is correct (edit: in degrees).

Your velocity in m/s looks wrong.

Your method also looks wrong. It makes no sense to get an acceleration by multiplying a velocity by a dimensionless number. Do you understand why this is wrong? vsin(theta) is meaningless here.

 

[tiny-tim]

A car is rising an excerpt in slope of 2%, at a constant speed of 80,5 km/h.
If the driver does not change the motion, or changing the pressure of your foot on the accelerator, which should be the acceleration of the car down the stretch of 3%?

Hi Apprentice123!

Hint: on the upslope, the acceleration is zero …

so what is the force per mass?

 

[cepheid]

I'm not asking you to resolve. I would like an explanation of the exercise

If the driver is moving at a constant speed, then a = 0. By Newton's second law, the net force on the car is zero. This means that

magnitude of forward force due to engine = magnitude of backward force due to weight. Therefore:

STEP 1: You can figure out how much force the engine is providing to drive the car upward.

Here, forward and backward are directions along the incline.

On the way down, the driver doesn't change the gas applied. Therefore, the force due to the engine is the same. This is the key to solving the problem.

 

[Apprentice123]

If the driver is moving at a constant speed, then a = 0. By Newton's second law, the net force on the car is zero. This means that

magnitude of forward force due to engine = magnitude of backward force due to weight. Therefore:

STEP 1: You can figure out how much force the engine is providing to drive the car upward.

Here, forward and backward are directions along the incline.

On the way down, the driver doesn't change the gas applied. Therefore, the force due to the engine is the same. This is the key to solving the problem.

I edited the speed (was wrong in the order of exercise).
Sorry, I do not understand English very well.
Are you saying that:

Psin(theta) = m.a ?
I find a = 0,294 m/s^2

 

[cepheid]

Psin(theta) = m.a ?
I find a = 0,294 m/s^2

Sorry, I don't know what P is.

Component of weight parallel to plane = ma

EDIT: I don't get the same answer as you.