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.(adsbygoogle = window.adsbygoogle || []).push({});

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.

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# Homework Help: Tangent Lines and Motion

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