A car rolls along the x-axis, and its position is shown as a function of time by

Question

A car rolls along the x-axis, and its position is shown as a function of time by the graph below. During which interval did the car have the higher average speed: t = [1, 3], or t = [1, 5] At which instant did the car have the higher Instantaneous speed: t = 2. or t = 4 ? What was the car's displacement during the interval t = [0, 9]? Which graph below best shows the car's instantaneous velocity as a function of time? The position y (in meters) of a yo-yo is given by the function , where t = time (in seconds). What is its displacement Delta y during the interval t =[0. 10) ? What is its average velocity during the same interval? What was its instantaneous velocity at 5 sec?

Explanation / Answer

(1) a. As you can see from the graph, the slope of the curve on t = [1,3] is mostly flat, so on average, the speed is very small. However, when you extend this to [1,5], you include a region about [3,5] where the car does have a nonzero speed. So, clearly, the higher average speed must be the one that includes a nonzero speed, or t = [1,5].

b. At t = 2, the curve is almost flat, whereas at t = 4 the curve has a large slope downwards. Since we are looking for the higher speed, clearly t = 4 is larger.

c. The car starts at x = 10m and ends at x = 0m, so the displacement is 10m.

d. The curve stars out flat, then slopes downwards, then flattens out again. So, clearly, the velocity starts out at zero, then becomes NEGATIVE (velocity, not speed, remember), then goes back to zero. Clearly, this is the first graph.

(2) a. y = y(10s) - y(0s) = (1 - 2*10 +0.5(10)^2 -0.03(10)^3) m - 1m = 0m

b. Average velocity is the displacement over the time interval, and since the displacement is zero, the average velocity is also 0.

c. dy/dt = v = - 2.0 + t - 0.09t^2, and

v(5s) = (- 2.0 + 5 - 0.09*(5)^2) m/s = 0.75 m/s

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