Answers to Selected Review Problems: Chapter 4

1. A(1) = 1; A(2) = 2.5; A(3) = 4,5; A(4) = 6; and A(5) = 7

3. The distance traveled is the area under the velocity curve.  There are 7.5 shaded squares in the figure, and each square is 10 feet/second × 10 seconds = 100 feet, so the car traveled 750 feet.

4. (a) Car A was traveling a constant 80 feet per second until t = 20 when its velocity started decreasing – it applied its brakes at t = 20.  It took another 20 seconds to stop.  Car B also applied its brakes at t = 20, but it took 40 seconds to stop.

(b) The distance the cars traveled after they started braking are represented by the area under the graph from t = 20 until they stopped.  For Car A, that area is 800 feet, so it traveled 800 feet before coming to a complete stop.  Car B also traveled 800 feet.

5. The police car will catch the speeder when the distance it has traveled is equal to the distance traveled by the speeder.  The area under the police car’s velocity after x seconds is 300 + 60(x-10); the area under the speeder’s velocity is 45x.  These are equal when x = 20.  The police car will catch the speeder after 20 seconds; they will each have traveled 900 feet.

7. You traveled 8 feet forward in the first 4 minutes, and then you traveled 8 feet backwards in the second 4 minutes.  You traveled a total of 16 feet, but you ended up the same place you started.  (In case you were wondering why you made so little progress?  You were practicing juggling while balancing on a tightrope.)