# Tangent Lines and Motion

1. May 25, 2006

### Ronnin

Well after a long gap between my last calculus course and my first physics course I will be taking this fall I have decided to do some brushing up. I am working though the MIT courseware for Cal I and have come to an example I am having trouble with.

How much must you slow down when a red light is 72 meters away? In 4 seconds it will be green. The waiting car will accelerate at 3 meters/sec^2. You cannot pass the car.

Strategy Slow down immediately to the speed V at which you will just catch that car. (If you wait and brake later, your speed will have to go below V.)At the catch-up time T, the cars have the same speed and same distance. Two conditions, so the distance functions in Figure 2.6d are tangent.
Solution At time T, the other car's speed is 3(T-4). That shows the delay of 4 seconds. Speeds are equal when 3(T-4) = V or T = 1/3 V +4. Now require equal distances. Your distance is V times T. The other car's distance is 72 + 1/2at^2:
72+1/2*3(T-4)^2=VT becomes 72+1/2*1/3V^2=V(1/3V+4).
The solution is V = 12 meters/second. This is 43 km/hr or 27 miles per hour. Without the other car, you only slow down to V= 72/4 = 18 meters/second. As the light turns green, you go through at 65 km/hr or 40 miles per hour.

The part I have trouble understanding is how the others car’s distance is 72+1/2at^2. Thanks in advance for any comments you can offer.

2. May 25, 2006

### HallsofIvy

Staff Emeritus
What about the "waiting car"?? The problem says that we must slow down when a red light is 72 meters away. What does that have to do with a waiting car?

Is it possible that the problem says: you are 72 meters behind a car waiting at a red light. After 4 seconds, the light will turn red and the car will accelerate at 3 m/s2. At what velocity must you decelerate in order to avoid hitting (or passing) the waiting car?

If so, that is quite different from the problem you posted!

And it would be simple if you were given your initial speed! Obviously the answer depends on how fast you are going to start with. Without that information you can't answer the question.

3. May 26, 2006

### Ronnin

HOI, thank you for your reply. I copied this problem exactly as it was written from Strang's Calculus Textbook. I believe you are correctly interpreting the question and it is just assumed that my car will instantaneously decelerate at the 72 meter mark to match a velocity and distance equivalent to the other car at some point after the 4 second wait. I still don’t see how the other car’s distance function is 72+1/2at^2? This problem is solved without doing any integration, but I tried to integrate twice back from the other car’s acceleration function and still can’t figure out how this distance was derived. Thank you so much for your time.

4. May 26, 2006

### HallsofIvy

Staff Emeritus
I don't see how it can be either! Does your solution book (or where ever you got that formula) tell what "t" means? That is, when is t= 0?
I see too possibilities: either t= 0 at the time you are 72 m from the light or t= 0 at the time the light turns green.

The "waiting car" doesn't start moving until the light turns green, 4 sec after your car is 72 m from the light. At the time the "waiting car" starts moving your car should be 72- (a/2)(4)2 from the light where you deceleration is a.

5. May 26, 2006

### Ronnin

Unfortunatly, the solutions manual doesn't cover the examples. Everything written is all that is given. I will do some more digging, but I think I will move on to some of the trig stuff and keep moving forward to reviewing some of the integration steps I have forgotten over the last 2 years. It doesn't take long to get rusty if you don't do this regularly. Thank you for all your help. At least I know I wasn't the only person confused by this problem.