Question

the table shows the motion data for two objects during two trials of an investigation. motion investigation data

object	trial	distance traveled (m)	time (s)
1	2	1,000	50
2	1	30	15
2	2	20	10

which object travels at the greater average speed?
a. object 1, because it travels a greater distance per unit of time.
b. object 1, because it travels the longest amount of time.
c. object 2, because it travels the shortest amount of time.
d. object 2, because it travels the shortest distance per unit of time.

Explanation:

Step1: Calculate average - speed formula

The formula for average speed is $v=\frac{d}{t}$, where $v$ is average speed, $d$ is distance traveled, and $t$ is time.

Step2: Calculate average speed of Object 1

For Object 1, in Trial 1, $d_1 = 500$ m and $t_1=40$ s, and in Trial 2, $d_2 = 1000$ m and $t_2 = 50$ s. The total distance $D_1=d_1 + d_2=500 + 1000=1500$ m, and the total time $T_1=t_1 + t_2=40 + 50 = 90$ s. The average speed $v_1=\frac{D_1}{T_1}=\frac{1500}{90}=\frac{50}{3}\approx16.67$ m/s.

Step3: Calculate average speed of Object 2

For Object 2, in Trial 1, $d_3 = 30$ m and $t_3=15$ s, and in Trial 2, $d_4 = 20$ m and $t_4 = 10$ s. The total distance $D_2=d_3 + d_4=30+20 = 50$ m, and the total time $T_2=t_3 + t_4=15 + 10=25$ s. The average speed $v_2=\frac{D_2}{T_2}=\frac{50}{25}=2$ m/s.

Step4: Compare average speeds

Since $\frac{50}{3}\approx16.67$ m/s $>2$ m/s, Object 1 has a greater average speed because it travels a greater distance per unit of time.

Answer:

A. Object 1, because it travels a greater distance per unit of time.