Physics 20: Uniform Acceleration Problem

[NHS]

Hello,

I'd first like to briefly introduce myself and say that any help you can lend is appreciated. I've been reading up here for quite some time and these forums are an incredible resource!

Here's the pickle:

An automobile is breaked to a stop with a uniform deceleration in a time of Ts. Show that the distance traveled during this time is given by d=Vo^2/a - 1/2aTs^2

I thought that it would have to involve combining two separate formulas to produce the one mentioned in the question. BUT, after much trying... and trying... and trying.. I can't seem to properly do it.

If anyone can give me some hints, tips (or even a solution) I'd be as happy as this guy --> Thanks!

 

[christinono]

Hello,

I'd first like to briefly introduce myself and say that any help you can lend is appreciated. I've been reading up here for quite some time and these forums are an incredible resource!

Here's the pickle:

An automobile is breaked to a stop with a uniform deceleration in a time of Ts. Show that the distance traveled during this time is given by d=Vo^2/a - 1/2aTs^2

I thought that it would have to involve combining two separate formulas to produce the one mentioned in the question. BUT, after much trying... and trying... and trying.. I can't seem to properly do it.

If anyone can give me some hints, tips (or even a solution) I'd be as happy as this guy --> Thanks!

I can't think of another way of showing that this is true other than using calculus.

 

[cepheid]

Actually, calculus is not required. As NHS originally thought, one needs only a basic knowledge of the kinematics of motion in a straight line with uniform acceleration. Some combination of two of the standard formulas will yield this result.

You're asked to show that the distance traveled by an object decelerating uniformly in a time t_s is given by this equation:

$$d = \frac{{v_0}^2}{a} - \frac{1}{2}a{t_s}^2$$

The key point in this problem is that the car is brought to a stop. So the final velocity v_f after accelerating for a time t_s is zero. From the standard formulas, we have:

$$0 = v_f = v_0 - at_s$$

Solve for t_s:

$$t_s = \frac{v_0}{a}$$

Consider the kinematics formula for the distance travelled:

$$d = v_{0} t_s - \frac{1}{2}a{t_s}^2$$

Substitute the above expression for t_s into the first term of the equation, and you get:

$$d = \frac{{v_0}^2}{a} - \frac{1}{2}a{t_s}^2$$

 

[NHS]

...and suddenly the veil of fog is lifted and I can see again!

Thanks so much. I understand now. I knew I was on the right path. *thumbs up*

 

[HallsofIvy]

Of course, you need caculus to show that those formulas are correct!

 

[cepheid]

Yes of course, I realize that, but that's not what the question is asking. When I took Physics 20, since it was my *first* ever course in physics (back in high school), the derivations of the kinematics formulas were presented to us using simple arguments e.g. find the area under the curve, which was easy to do for constant a. If you're still around NHS, I'm curious: are you from Alberta? Because I always thought our naming system there was peculiar (gr 11 courses were "20" level courses, and gr. 12 courses were "30" level courses. Why not just physics 11 and physics 12 instead? I'd be really surprised if somebody else had these names too, unless it's some other Canadian province.