Question

1. a car travels 20 meters east in 1.0 second. the displacement of the car at the end of this 1.0 - second interval is

a. 20 m
b. 20 m/s
c. 20 m east
d. 20 m/s east

1. a high - speed train in japan travels a distance of 300. kilometers in 3.60×10³ seconds. what is the average speed of this train?

a. 1.20×10⁻² m/s
b. 8.33×10⁻² m/s
c. 12.0 m/s
d. 83.3 m/s

1. what is the distance traveled by an object that moves with an average speed of 6.0 meters per second for 8.0 seconds?

a. 0.75 m
b. 1.3 m
c. 14 m
d. 48 m

1. a girl leaves a history classroom and walks 10. meters north to a drinking fountain. then she turns and walks 30. meters south to an art classroom. what is the girls total displacement from the history classroom to the art classroom?

a. 20. m south
b. 20. m north
c. 40. m south
d. 40. m north

1. one car travels 40. meters due east in 5.0 seconds, and a second car travels 64 meters due west in 8.0 seconds. during their periods of travel, the cars definitely had the same

a. average velocity
b. total displacement
c. change in momentum
d. average speed

1. a car travels between the 100 - meter and 250 - meter highway markers in 10 seconds. the average speed of the car during this interval is

a. 10 m/s
b. 15 m/s
c. 25 m/s
d. 35 m/s

1. what is the average velocity of a car that travels 30 kilometers due west in 0.50 hour?

a. 15 km/hr
b. 60 km/hr
c. 15 km/hr west
d. 60 km/hr west

1. an athlete can run 9 kilometers in 1 hour. if the athlete runs at that same average speed for 30 minutes, how far will the athlete travel?

a. 18 kilometers
b. 9 kilometers
c. 4.5 kilometers
d. 3.3 kilometers

Explanation:

Step1: Recall displacement definition

Displacement is a vector quantity with magnitude and direction. A car travels 20 meters east in 1.0 second, so displacement is 20 m east.

Step1: Use average - speed formula

The formula for average speed is $v=\frac{d}{t}$. Given $d = 300\times10^{3}$ m and $t=3.60\times 10^{3}$ s. Then $v=\frac{300\times 10^{3}}{3.60\times 10^{3}}=\frac{300}{3.60}\approx83.3$ m/s.

Step1: Apply distance - speed - time relation

The formula for distance is $d = v\times t$. Given $v = 6.0$ m/s and $t = 8.0$ s. So $d=6.0\times8.0 = 48$ m.

Step1: Calculate net displacement

She moves 10 m north and 30 m south. Net displacement $d=10 - 30=- 20$ m (negative means south direction).

Answer:

C. 20 m east