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Question
- a baseball pitcher throws a fastball at 42 meters per second. if the batter is 18 meters 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 10. a ball is dropped from rest from a height 6.0 meters above the ground. the ball falls freely and reaches the ground 1.1 seconds later. what is the average speed of the ball? a. 5.5m/s b. 6.1m/s c. 6.6m/s d. 11m/s 11. imagine that you are sitting in the front seat of a bus that is traveling down a straight highway at a speed of 28 meters/second. you roll a ball down the aisle to your friend who is sitting in the back of the bus. if the ball rolls with a constant speed of 1 meter/second away from you, how fast is the ball traveling relative to the highway? a. 1 meter/second b. 27 meters/second c. 28 meters/second d. 29 meters/second 12. distance is to displacement as a. force is to weight b. speed is to velocity c. velocity is to acceleration d. impulse is to momentum 13. the speedometer in a car does not measure the cars velocity because velocity is a a. vector quantity and has a direction associated with it b. vector quantity and does not have a direction associated with it c. scalar quantity and has a direction associated with it d. scalar quantity and does not have a direction associated with it 14. a softball player leaves the batters box, overruns first base by 3.0 meters, and then returns to first base. compared to the total distance traveled by the player, the magnitude of the players total displacement from the batters box is a. smaller b. larger c. the same 15. a baseball player runs 27.4 meters from the batters box to first base, overruns first base by 3.0 meters, and then returns to first base. compared to the total distance traveled by the player, the magnitude of the players total displacement from the batters box is a. 3.0m shorter b. 6.0m shorter c. 3.0m longer d. 6.0m longer 16. a car travels 300 meters in 15 seconds. what is the average velocity of the car? a. 5.0 m/sec b. 15 m/sec c. 20 m/sec d. 30 m/sec
9.
Step1: Recall speed - distance - time formula
We know that speed $v=\frac{d}{t}$, and we want to find time $t$. Rearranging the formula gives $t = \frac{d}{v}$.
Step2: Substitute values
Given $d = 18$ meters and $v=42$ m/s. Then $t=\frac{18}{42}\approx0.43$ s.
Step1: Recall average - speed formula
The average - speed formula is $v_{avg}=\frac{d}{t}$.
Step2: Identify values
The distance $d = 6.0$ meters and the time $t = 1.1$ seconds. Then $v_{avg}=\frac{6.0}{1.1}\approx5.5$ m/s.
Step1: Analyze relative - motion
The speed of the bus is $v_{bus}=28$ m/s and the speed of the ball relative to the bus is $v_{ball - bus}=1$ m/s in the same direction as the bus. The speed of the ball relative to the highway is the sum of the speed of the bus and the speed of the ball relative to the bus.
Step2: Calculate relative speed
$v = v_{bus}+v_{ball - bus}=28 + 1=29$ m/s.
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