Calculating for a distance including vi, vf, time, and Ff

In summary, the conversation discusses using variables to calculate the distance and acceleration of a DeLorean in a scene from the movie Back to the Future. The given variables include time, initial and final velocities, coefficient of friction, and mass of the car. The attempt at a solution involves calculating the distance and friction separately and considering the frictional force in the calculation of distance. However, it is mentioned that the friction calculation may not be correct and the frictional force does not need to be considered in the distance calculation.
  • #1
varoney
1
0

Homework Statement


For a project we have to find a movie scene and calculate using any variables we can find. I am using Back to the Future and want to calculate the distance the DeLorean travels in one of the first scenes when they're first using it. I'd also like to find the acceleration from the same scene.

Homework Equations


I know:
t= 12.35s (From the time the car begins to move to when it reaches 88 mi/h)
d=? a=?
vi= 28.07411 m/s (The speed the car reaches when he releases it)
vf= 39.3395 m/s
μ= 0.72 (car tires on asphalt)
m=1,230 kg
I think I calculated properly for Ff as well which comes to 8687.736 N

The Attempt at a Solution


I have calculated separately for distance and for friction but I don't know if my friction calculation is correct or how to apply the number to the calculation for distance.
 
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  • #2
varoney said:

Homework Statement


For a project we have to find a movie scene and calculate using any variables we can find. I am using Back to the Future and want to calculate the distance the DeLorean travels in one of the first scenes when they're first using it. I'd also like to find the acceleration from the same scene.

Homework Equations


I know:
t= 12.35s (From the time the car begins to move to when it reaches 88 mi/h)
d=? a=?
vi= 28.07411 m/s (The speed the car reaches when he releases it)
vf= 39.3395 m/s
μ= 0.72 (car tires on asphalt)
m=1,230 kg
I think I calculated properly for Ff as well which comes to 8687.736 N

The Attempt at a Solution


I have calculated separately for distance and for friction but I don't know if my friction calculation is correct or how to apply the number to the calculation for distance.
Here V(f), V(i) and time are given, by these data you can simply calcute net acceleration which is equal to acceleration provided by engine to car minus acceleration provide by frictional force.so there is no need to consider this number in calculation of distance
 

1. How do I calculate distance with initial and final velocities, time, and final force?

To calculate distance, you can use the formula d = (vi + vf)t/2 + Ff*t^2/2, where vi is the initial velocity, vf is the final velocity, t is the time, and Ff is the final force. Simply plug in the values and solve for distance.

2. Can I calculate for distance if I only have initial and final velocities?

Yes, you can use the formula d = (vi + vf)t/2, where vi is the initial velocity, vf is the final velocity, and t is the time. This formula assumes that there is no final force acting on the object.

3. What if I only have the initial velocity and time?

In this case, you can use the formula d = vit + at^2/2, where vi is the initial velocity, a is the acceleration, and t is the time. This formula assumes that the final velocity is 0.

4. How do I calculate for distance if I have varying forces acting on the object?

If you have varying forces acting on the object, you can use the equation d = (vi + vf)t/2 + ΣF*t^2/2, where vi is the initial velocity, vf is the final velocity, t is the time, and ΣF is the sum of all the forces acting on the object.

5. Is there a simpler way to calculate distance with multiple variables?

Yes, you can use the formula d = vit + (1/2)(vf - vi)t, which combines the initial and final velocities with the time to calculate distance. This formula assumes that there is no final force acting on the object.

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