A computer-controlled racecar is programmed to execute thefollowing motion along the ground for 6.0 seconds. Let’s say thecar begins at the origin of our coordinate system. Its initialvelocity is ~v0 = (15:0 m/s)^i and its acceleration is constant: ~a= (?6:0 m/s2)^i + (?2:0 m/s2)^j

(a) Make a table, and calculate the car’s position vector, ~r atthe end of each second, through t = 6:0 seconds. Use these data toplot the trajectory of the particle for this time interval. You cando it by hand on the axes given to you on the last page. (If youlike, you could also do this by writing a computer program thatwould calculate hundreds of data points).

(b) On your graph, sketch vectors for both the car’s velocityand acceleration at t = 1:0; 2:0 and 4.0 seconds. At each of theseinstants, is the car speeding up or slowing down? How do youknow?

(c) You may have found that judging the answer to part (b) was atough call for t = 2:0 s. We can do this conclusively: i. Find boththe velocity and the speed of the particle at t = 2:0 seconds. ii.At t = 2:0 seconds, find the rate at which the car’s speed ischanging (Note that I’m not asking the rate at which its velocityis changing, which is just j~aj). (Hint: you’ll need to find theangle between ~v and ~a, which is something for which the vectordot product is very useful... ) iii. At what time is the rate ofchange of the car’s speed equal to zero, and what is the car’sspeed at this instant? 3