Physics question?

Question

If you were decelerating, but you doubled your initial speed...what does this do to the time and distance it takes to stop my car?

To do this would I just plug in random numbers?


I know i'm supposed to double the time but I don't understand WHY!!!

Answer 1

Didn't you already asked this questions? Still don't understand?
As other have said, you don't need pluging in any numbers, this is about understanding the relationship between velocity, time and distance.

So given your scenario of a car moving at a constant velocity, V, we want to apply the brakes (at a constant deceleration) and come to a complete stop. Now it turns out that it doesn't matter what the deceleration is, as long as it is constant (meaning that you applied the brake evenly the whole time for the car to come to a stop).

So for some given velocity, V (and it also doesn't matter what it is as long as it is constant), it will take some amount of time, T, for the car to stop. The equation for finding the time, T is:

V(final) = V(initial) - aT, and since V(final) of the car is 0, because it has come to a complete stop,

0 = V(initial) - aT => T = V(initial)/a

Now we have the relationship between initial velocity and time, T is directly proportional to V(initial). So if we were to double V(initial), we then also double the time, T.

If V(initial) = 4 m/s, and a = 2 m/s^2 then T=4/2 = 2 sec

But if V(initial) is 8 m/s or double the original speed,
then T = 4 sec. Hence doubling the initial velocity doubles the time it takes to stop the car completely.

Similarly for understanding the relationship between initial velocity and the braking distance, or the distance the car will continue to travel before coming to a complete stop, we have the following equation:

D = (1/2)aT^2 describes the relationship between distance and time. Since, we already know from before that T = V(initial)/a, we plug that into the equation and get:

D = (1/2)a(V(initial)/a)^2 = (1/2a)V(initial)^2

So notice that the relationship between D, the distance, and V the initial velocity is a "square". This means that if we double the initial velocity, we will quadruple (4 times) the distance!

If V(initial) = V, then D=(1/2a)V^2

If V(initial) = 2V, then D=(1/2a)(2V)^2 = (1/2a) 4V^2 = (2/a)V^2

Notice that (2/a)V^2 is 4 times bigger than (1/2a)V^2.

Now we plug in some numbers to see how this works:

D = (1/2a)V(initial)^2

a=2 m/s^2 and V(initial)=4 m/s, we get D = 4 m
a=2 m/s^2 and V(initial)=8 m/s, we get D = 16 m

So the braking distance is 4 times longer if we double the initial velocity!

Answer 2

Better than plugging random numbers into a formula is to understand WHY it takes twice as long (in seconds) and four times as far (in feet) to stop your car.

Here goes:
When you execute a maximum stop (i.e., using all of your coefficient of friction with the road, but without skidding*), your tires exert a stopping force equal to the car's weight times the coefficient of static friction between the tire and the road. (The tires exert this force against the road, and the road exerts this force against the tire, and that force against the tire causes the car to slow down.)

Because the car's weight does not change (significantly) during the stop, and the coefficient of friction does not change, the stopping force during this maximum stop is constant. (You can argue with those assumptions, because the car's weight shifts toward the front tires, the road surface may vary and cause the coefficient of friction to vary, etc., etc., but reality comes pretty close to doing what the preceding sentence says.)

So what does a constant stopping force mean for the time required to stop? Since F = ma, a constant F (force) and a constant m (mass of the car) mean a constant a (acceleration; which means deceleration in this case).

And if your deceleration is constant, you can just divide your speed (e.g., 30 mph) by this constant deceleration (e.g., 10 mph per second) to find the time required to stop (3 seconds). If you were initially going 60 mph, you'll have to divide 60 by 10 yourself; I can't do all the work for you.

But WHY is the distance you traveled while stopping FOUR times as far? Simple. The time to stop is twice as long, but the AVERAGE SPEED is also twice as long. Twice the speed for twice as many seconds means four times the distance.

* If you skid the tires, it will take longer (and more distance) to stop because the stopping force will be less. It will equal the weight of the car times the coefficient of SLIDING friction (which is less than the coefficient of static friction). So DON'T SKID when you stop. If you have ABS, the car takes care of it for you. If not, you can try to find the brake pressure that maximizes your stopping force without skidding, or you can pump the brakes (to make sure that at least PART of the time you're not skidding).

Drive safely!

Answer 3

it takes x amount of time for the car to come to a complete stop.
when the velocity is doubled and the car is still decelerating at the same rate it takes double the time for the car to come to a complete stop.