Responses:

1. Velocity, in the most formal sense, is the rate of change of position. Think of position as a point on the coordinate plane (x,y). Think of the new position exactly 1 second later as (x₁, y₁). You can use the Pythagorean theorem to determine the distance between the two points (try it with the points (3,5) and (7,8). You should calculate a distance of 5 units. Since the position changed by 5 units in 1 second we can say the velocity is 5 units/second (or 5 m/s if the “unit” is the meter). Note that how we got from (3,5) to (7,8) doesn’t matter – a curve or a straight line represents the same change in position. If we were worried about speed we’d have to consider the length of the path from the first point to the other – and the speed would be greater than the velocity for any other than a straight line path.

2. This requires us to assume straight line motion. Velocity is the rate of change of position and can be calculated by dividing the displacement (240 km) by the time (4 hours).

a. 240 km ÷ 4 hrs = 60 km/hr.

b. 240 km = > 240 km • 1000 m/km = 240,000 m

c. 4 hrs => 4 hr • 3600s/hr = 14,4000 s

d. 240000 m ÷ 14400 s = 16.7 m/s

3. Acceleration is the rate of change of velocity. If an object (such as the car in the previous problem) moves at constant velocity, its acceleration is zero. If the velocity is changing, the amount of the change (∆V ) is divided by the amount of time required to bring about that change. The conventional units of acceleration are m/s/s (read as meters per second, per second). It sounds funny at first, but think of it as by how many meters per second the velocity must change each (per) second. The sign of acceleration is negative if an object is slowing down in the positive direction or speeding up in the negative direction.

4. The change in velocity is:

a. 30 m/s – 0 m/s = 30 m/s.

b. The time required is 10 seconds.

c. The acceleration is 30 m/s ÷ 10 s = 3 m/s/s

5. The distance is the product of average velocity and time:

a. For uniform acceleration (the only kind in this class) the average velocity is the average of the initial velocity and the final velocity. Here: (0 + 30) ÷ 2 = 15 m/s

b. So the car moves at an average of 15 m/s for 10 seconds. 15 m/s • 10 s = 150 meters

6. At 2.0 seconds the circle is at the 2 meter mark. At 3.0 seconds at the 4 meter mark. The change in position is 4 – 2 = 2 meters. The time is 1 second. Velocity = 2 m ÷ 1 s = 2 m/s. Or try: at 4.0 seconds the circle is at the 6.0 meter mark. Change in position is 6 – 2 = 4 meters. Time is now 2 seconds. 4 meters ÷ 2 seconds = 2 m/s.

7. The ball moves 3 meters in the first 2 seconds. An average of 1.5 m/s at the 1 second mark. The ball moves 9 meters during the next 2 seconds. An average of 4.5 m/s at the 3 second mark. a = ∆V/t = (4.5 – 1.5) m/s ÷(3 – 1) s = 1.5 m/s/s.

8. Your printer might get different results, or you might measure in inches. In any event the acceleration you get should be nearly the same as mine. I get the following values in millimeters from left to right: 4, 9, 13, 17, 22, 27, 31, 35, 40. If 4 mm corresponds to 3 meters on our scale, I can say we have 0.75 m/mm. In the first two intervals I have 13 mm •0.75 m/mm =9.75 meters over 2/60^th of a second gives 292.5 m/s of velocity at the second dot. Between the 6^th and 8^th I have 31 + 35 = 66 mm. 66 • 0.75 m/mm = 49.5 meters in 2/60^th second (gives 1485 m/s at the seventh dot). The second dot is 5/60^th or 1/12^th of a second before the seventh. The change in velocity is (1485 -292.5 = 1192.5 m/s) over 1/12^th second = 14,310 m/s/s.