Motion at constant acceleration problem

In summary: So you havev^2 = vi^2 + 2a(x-xi)You know x-xi, you know a, and you know v. Solve for vi.One note: in this problem, you can also use x = x0 + (v0)t + (1/2)at^2, since you have x, x0, a, and v0. In summary, in this problem, a car traveling at an unknown speed leaves skid marks 92 m long due to a constant deceleration of 7.00 m/s^2. Given the final position, acceleration, and final velocity, the car's initial velocity can be determined using the equation v^2
  • #1
cubejunkies
34
0
In coming to a stop, a car leaves skid marks 92 m long on the highway. Assuming a constant deceleration of 7.00 m/s^2, estimate the speed of the car just before breaking.

I think my knowns are the initial position (0 m), final position (92m), acceleration (7.00 m/s^2) and final velocity (0 m/s). But I don't have the time it took to slow down and idk, none of the equations I have for motion at constant acceleration really look like they would help me solve the problem. I apologize for this silly question because I just learned about motion at constant acceleration a few days ago and so I'm not really good at this at all :/

Thank you so much for helping!
 
Physics news on Phys.org
  • #2
cubejunkies said:
In coming to a stop, a car leaves skid marks 92 m long on the highway. Assuming a constant deceleration of 7.00 m/s^2, estimate the speed of the car just before breaking.

I think my knowns are the initial position (0 m), final position (92m), acceleration (7.00 m/s^2) and final velocity (0 m/s). But I don't have the time it took to slow down and idk, none of the equations I have for motion at constant acceleration really look like they would help me solve the problem. I apologize for this silly question because I just learned about motion at constant acceleration a few days ago and so I'm not really good at this at all :/

Thank you so much for helping!
Other than the initial position, three pieces of information are always sufficient to solve a (one-dimensional) constant acceleration problem. Since you have three pieces of information -- final position, acceleration, and final velocity -- this is solvable.

Try having another look at those equations. You want to find one that contains final position, acceleration, and final velocity in it.

p.s. Remember to watch the +/- signs. Are you sure the acceleration is what you said it is? :wink:
 
  • #3
cubejunkies said:
none of the equations I have for motion at constant acceleration really look like they would help me solve the problem.

What equations you have for constant acceleration? You have to use them, just think of how.
 
  • #4
Would the acceleration be considered negative? If the car is slowing down, its acceleration is not always necessarily negative, but would this be a situation where it is?

The equations I have:

v= vi + at
x= xi + vi*t + (1/2)(a)(t)^2
v^2 = vi^2 + 2a(x-xi)
 
Last edited:
  • #5
cubejunkies said:
Would the acceleration be considered negative? If the car is slowing down, its acceleration is not always necessarily negative, but would this be a situation where it is?
The idea is to pick a direction to be positive, and be consistent. If you pick the direction the car is traveling to be positive, then yes the acceleration is negative.

The equations I have:

v= vi + at
x= xi + vi*t + (1/2)(a)(t)^2
v^2 = vi^2 + 2a(x-xi)
Good. Which one of those contains displacement, final velocity, and acceleration?
 
  • #6
Redbelly98 said:
The idea is to pick a direction to be positive, and be consistent. If you pick the direction the car is traveling to be positive, then yes the acceleration is negative.


Good. Which one of those contains displacement, final velocity, and acceleration?

Would it be the last one? I think that taking the acceleration to be negative would help, I tried this one before but got some outlandish number :P
 
  • #7
cubejunkies said:
Would it be the last one? I think that taking the acceleration to be negative would help, I tried this one before but got some outlandish number :P

That outlandish number wouldn't have been equivalent to 2881 miles per hour would it?
 
  • #8
cubejunkies said:
Would it be the last one?
Yes.
 

Related to Motion at constant acceleration problem

1. What is motion at constant acceleration?

Motion at constant acceleration refers to the movement of an object at a consistent rate of change in velocity. This means that the velocity of the object is changing by the same amount over equal intervals of time.

2. How is acceleration calculated in a constant acceleration problem?

Acceleration can be calculated by dividing the change in velocity by the change in time. This is represented by the equation a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

3. What is the difference between speed and velocity in a constant acceleration problem?

Speed refers to the rate of motion, while velocity refers to the rate of motion in a specific direction. In a constant acceleration problem, the speed of an object may be changing, but the velocity remains constant if the direction of motion does not change.

4. How is distance calculated in a constant acceleration problem?

Distance can be calculated by using the equation d = vit + 1/2at^2, where d is distance, vi is initial velocity, a is acceleration, and t is time. This equation is derived from the definition of acceleration and the formula for calculating average velocity.

5. What are some real-life examples of motion at constant acceleration?

Some examples of motion at constant acceleration include a car traveling at a constant speed on a straight road, a ball rolling down an inclined plane, and a rocket taking off from the ground. In all of these cases, the acceleration is constant because the object is moving in a straight line with a consistent change in velocity.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
310
  • Introductory Physics Homework Help
2
Replies
35
Views
2K
  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
2
Replies
38
Views
3K
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
20
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
Back
Top