Question

brian, kelsey, and geoff each have a remote - controlled car. they simultaneously started their cars and drove them in a straight line away from a motion sensor. the sensor recorded the distance to each car in the table below.

distance from sensor to each car

time since starting cars	distance to brians car	distance to kelseys car	distance to geoffs car
3 sec	38 cm	31 cm	33 cm
5 sec	42 cm	35 cm	39 cm
8 sec	48 cm	41 cm	48 cm

which statement is best supported by the table?
o brians and kelseys cars traveled at the same rate.
o kelseys and geoffs cars traveled at the same rate.
o brians and geoffs cars started the same distance from the sensor.
o kelseys and geoffs cars started the same distance from the sensor.

Explanation:

Step1: Calculate rate of change for Brian

Rate = $\frac{\Delta d}{\Delta t}$, for Brian from $t = 1$ to $t = 3$: $\frac{38 - 34}{3 - 1}=\frac{4}{2}=2$ cm/s.

Step2: Calculate rate of change for Kelsey

For Kelsey from $t = 1$ to $t = 3$: $\frac{31 - 27}{3 - 1}=\frac{4}{2}=2$ cm/s. But from $t=3$ to $t = 5$, Brian's rate is $\frac{42 - 38}{5 - 3}=2$ cm/s and Kelsey's is $\frac{35 - 31}{5 - 3}=2$ cm/s. Overall, we need to check all intervals. For Brian from $t = 5$ to $t = 8$: $\frac{48 - 42}{8 - 5}=2$ cm/s, for Kelsey from $t = 5$ to $t = 8$: $\frac{41 - 35}{8 - 5}=2$ cm/s. They have same rate, but let's check others.

Step3: Calculate rate of change for Geoff

For Geoff from $t = 1$ to $t = 3$: $\frac{33 - 27}{3 - 1}=\frac{6}{2}=3$ cm/s. So Brian and Kelsey don't have same rate as Geoff.

Step4: Check starting - distance

At $t = 1$ s, distance of Kelsey's car is 27 cm and distance of Geoff's car is 27 cm. So Kelsey's and Geoff's cars started the same distance from the sensor.

Answer:

Kelsey's and Geoff's cars started the same distance from the sensor.