What is the Maximum Speed for Avoiding a Barrier with Constant Acceleration?

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The discussion revolves around calculating the maximum speed a car can travel without hitting a barrier 36 meters ahead, given a reaction time of 0.95 seconds and a braking acceleration of -10.0 m/s². Participants clarify that the initial speed (v) is what needs to be determined, while the provided speed of 100 km/h is irrelevant for the calculation. The problem is divided into two parts: the distance covered before braking and the distance during braking. Key equations of motion are suggested to relate initial velocity, final velocity, acceleration, and distance. Ultimately, the focus is on using the correct equations to solve for the maximum safe speed without collision.
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Hello all, new to the forums.. I am absolutely stuck on this problem for my physics tonight. It's dealing with acceleration, so here it is:

A driver of a car going 100.0 km/h suddenly sees the lights of a barrier 36.0 m ahead. It takes the driver 0.95 s before he applies the brakes, and the average acceleration during braking is -10.0 m/s^2.
What is the maximum speed at which the car could be moving and not hit the barrier 36.0 m ahead? Assume the acceleration rate doesn't change.

Your help is appreciated!
 
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Well, I don't see what the 100 km/hr has to do with the problem since you're asked for the maximum speed it could be travelling, so I guess we'll ignore it.

The problem exists in two parts - the part before the driver applies the brakes and the part while the brakes are on. Let the speed of the car be v. In the first part, you know the speed is constant, so the distance traveled is .95v (assuming v to be in meters per second.) For the second part of the problem, you have the initial velocity (v), the acceleration (-10 m/s^2), and if we assume that v is the maximum possible safe speed, we have the final speed, 0. We also have the total distance traveled - 36m. Note that this distance includes both the distance traveled in the first part as well as that traveled in the second part.

So, you need a relation between initial velocity, final velocity, acceleration and distance. Add that to what you get from the first part, set them equal to 36 m, and solve for v.

Make sense?
 
Heh.. trying to make sense out of it.. however, I'm still a little confused on the second part.. I'm not quite sure what the initial velocity is however. I understand the total distance traveled is 36m. I'm just confused about v.
 
The initial velocity in the second part is still v, because the velocity doesn't change in the first part. No brakes.
 
so what is v, 100km/h?
 
No, hon. v is what you're trying to find. You have all the other information. As I said, I have no clue in the world what the 100 kph is doing there.
 
Oh.. ok, what equation can I use to figure this out?
 
You're asking too much there, honey. Go back to your book and look at the equations of motion you have. There's probably already one sitting there with the quantities you need. If there isn't, I guarantee there are two of them that you can put together to get what you need.

Patience is a virtue. :)
 
ok, i have two equations here that will get me final velocity.. I don't have any that would get me initial velocity. The two I have are V_f = V_i +at ... and ... V_f^2 = V_i^2 + 2ad.. do I use one of these? and if I do.. I don't know the value of V_i.. a hint.. or anything actually helpful would be nice.
 
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Again - the initial velocity is what you're looking for. You want a relation between the final velocity (0), the acceleration, the distance, and the initial velocity. How about

vf^2 - v0^2 = 2ad
 
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