Ch 1.3

An automobile travels between the 100-m and 250-m markers on a highway in 10s. The average speed of the automobile during this interval is a)10m/s b)15m/s c)25m/s d)40m/s

B-Average speed is distance traveled/time needed to go that distance. The distance between the markers was 150m(250m-150m), so the speed was 150m/10s or 15m/s

The pattern shown below was left by an automobile as it dropped oil at a constant rate on the road while being driven. During which section of the trip was the car traveling with constant velocity? a)AB b)BC c)CD d)DE

B-One of the definitions of constant velocity is equal distances in equal times. Because the oil comes out a constant rate, the area where the dots are equally spaced is where the speed will be constant. In section AB, the distance between the dots is increasing, indicating the car was accelerating. CD and DE show non-uniform motion with dots being unevenly spaced.

The graph represents the relationship between distance and time for an object in straight-line motion. During which interval was the object at rest? a)AB only b)BC only c)CD only d)both AB and CD

D-During intervals AB and CD, the automobile was the same distance during the time interval, so it's speed must be zero. During interval 0A, the automobile was moving away from the start, and during interval BC, it was moving back toward the starting point.

The graph shows distance traveled by two objects, A and B. Compared to the speed of object B, the speed of object A is a)the same b)twice as great c)one half as great d)three times as great

D-The slope of A is 4 times the slope of B, so the speed must be 3 times greater. To get the slope, the students can divide the change in the distance by the change in time. Slope A has a change of 60m in the same time as slope B's change of 20m.

A baseball pitcher throws a fastball at 42 m/s. If the batter is 18 m from the pitcher, approximately how much time does it take for the ball to reach the batter? a)1.9s b)2.3s c)0.86s d)0.43s

D-v=d/t, where v=42m/s, d=18m. 42m/s=10m/t. Solving for t gives 0.43s.