Motion at constant acceleration problem

AI Thread Summary
To estimate the speed of a car just before braking, given a skid mark length of 92 meters and a constant deceleration of 7.00 m/s², the relevant equation is v² = vi² + 2a(x - xi). The initial position is 0 m, the final position is 92 m, and the final velocity is 0 m/s, indicating the car is coming to a stop. The acceleration should be considered negative since the car is decelerating. Using the appropriate equation with these values allows for the calculation of the initial speed before braking.
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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!
 
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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:
 
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.
 
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)
 
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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?
 
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
 
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?
 
cubejunkies said:
Would it be the last one?
Yes.
 
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