Question

1. mr. harper ran a 200 meter dash in 26.6 seconds. what is his average speed and average velocity? (s=d/t)(v=d/t)
2. mr. harper decided to quit teaching to play professional baseball. he can throw a fastball that travels 4.9 meters in 0.13 seconds. what is the average speed and velocity of his fastball? (s=d/t)
3. a car travels from point a to point b in 3 hours and back to point a in 5 hours. point a and point b are 157km apart. what is the total displacement of the car and what was its average velocity? (v=d/t) (displacement=final position - initial position)

Explanation:

Step1: Calculate average speed for Mr. Harper's 200 - meter dash

For average speed $S$, using the formula $S = \frac{d}{t}$, where $d = 200$ meters and $t=26.6$ seconds. So $S=\frac{200}{26.6}\approx7.52$ m/s. Since the 200 - meter dash is a straight - line motion in one direction, the magnitude of average velocity $V$ is the same as average speed in this case, $V = 7.52$ m/s.

Step2: Calculate average speed for Mr. Harper's fastball

Using the formula $S=\frac{d}{t}$, with $d = 4.9$ meters and $t = 0.13$ seconds. Then $S=\frac{4.9}{0.13}\approx37.69$ m/s. Assuming the fastball moves in a straight - line (neglecting any curve for simplicity), the magnitude of average velocity is the same as average speed, $V = 37.69$ m/s.

Step3: Calculate displacement and average velocity for the car

The car travels from point A to point B and then back to point A. Displacement $D$ is the final position minus the initial position. Since the car returns to its starting point, $D=0$ km. Using the formula $V=\frac{D}{t}$, and the total time $t=3 + 5=8$ hours. So $V=\frac{0}{8}=0$ km/h.

Answer:

1. Average speed: $7.52$ m/s, Average velocity: $7.52$ m/s
2. Average speed: $37.69$ m/s, Average velocity: $37.69$ m/s
3. Displacement: $0$ km, Average velocity: $0$ km/h