1. I have a simple question about acceleration & relative distance. Please excuse the inelegant phrasing, I'm not a scientist.

Driving in a semi-rural area, I come to a stoplight at a 4-way intersection. When the signal is in my favor, I turn left, (heading northward), and start to accelerate to the posted speed limit of 50 mph. Shortly after I complete my turn, the signal changes, and a waiting car that is also heading north begins to accelerate, behind me. At the moment the other motorist begins moving, I'm 100 yards ahead of him, and my speed is 25 mph.

The other motorist accelerates at twice the rate that I do, so that we both reach a speed of 50 mph at the same time. At that moment, is he still 100 yards behind me? Or less? Or more?  2.

3. As long as you are moving faster than him, you are opening up your lead. When you are both going the same speed, you will maintain the same separation.

I suppose now you want to know how far ahead you'll be? Well, you can use s=v0t+(1/2)at^2 to figure it out, if you knew t, i.e., how long it took you both to get up to 50 mph. Then a would be the change in velocity divided by t assuming constant acceleration.  4. Originally Posted by Harold14370 As long as you are moving faster than him, you are opening up your lead. When you are both going the same speed, you will maintain the same separation.

I suppose now you want to know how far ahead you'll be? Well, you can use s=v0t+(1/2)at^2 to figure it out, if you knew t, i.e., how long it took you both to get up to 50 mph. Then a would be the change in velocity divided by t assuming constant acceleration.

That's what I was thinking, but I wasn't 100% sure.

Thanks.  5. Let's put some numbers to it. If you had constant acceleration in the first hundred yards, then s1=100 = (1/2)a(t1)^2. If s2 is the distance after you have reached 50 mph, then the time is 2t1 (because it took you t1 seconds to get from zero to 25 mph, it will take you 2t1 to get from zero to 50 mph) and s2=(1/2)a(2t1)^2 and s2/s1=4, so your total distance traveled past the intersection 400 yards. For the other guy, his acceleration is twice yours, so it's 2a, and his time of travel is t1, so s3=(1/2)2a(t1)^2
s3/s2=2a/4a, and S3=s2/2.

He has gone half as far as you, or 200 yards, and so you have gone from 100 yards ahead to 200 yards ahead.  Bookmarks
 Posting Permissions
 You may not post new threads You may not post replies You may not post attachments You may not edit your posts   BB code is On Smilies are On [IMG] code is On [VIDEO] code is On HTML code is Off Trackbacks are Off Pingbacks are Off Refbacks are On Terms of Use Agreement