Question

nancy and jerome conducted an investigation using a toy car. in the table, they recorded the distance and time that the car traveled
toy car investigation data
distance(cm) time (sec)
0 0
10 2
20 4
30 5
40 8
50 10
which conclusion about the motion of the toy car is supported by the data in the table?
the car traveled at a constant speed throughout the entire distance.
the car accelerated during each measured interval.
the car was accelerating from 20 - 30 cm.
the car was decelerating from 40 - 50 cm.

Explanation:

Step1: Calculate speed for each interval

Speed formula: \(v=\frac{d}{t}\)
For \(0 - 10\) cm: \(v_1=\frac{10}{2} = 5\) cm/s
For \(10 - 20\) cm: \(v_2=\frac{20 - 10}{4 - 2}=\frac{10}{2}=5\) cm/s
For \(20 - 30\) cm: \(v_3=\frac{30 - 20}{5 - 4}=\frac{10}{1}=10\) cm/s
For \(30 - 40\) cm: \(v_4=\frac{40 - 30}{8 - 5}=\frac{10}{3}\approx3.33\) cm/s
For \(40 - 50\) cm: \(v_5=\frac{50 - 40}{10 - 8}=\frac{10}{2}=5\) cm/s

Step2: Analyze each option

• Option 1: Speeds \(v_1 = 5\), \(v_2 = 5\), \(v_3 = 10\), \(v_4\approx3.33\), \(v_5 = 5\) are not constant. So, this option is wrong.
• Option 2: Since speeds are not increasing in each interval (e.g., \(v_3 = 10\), \(v_4\approx3.33
• Option 3: \(v_2 = 5\) (speed before \(20 - 30\) cm interval), \(v_3 = 10\) (speed in \(20 - 30\) cm interval). As \(v_3>v_2\), the car was accelerating from \(20 - 30\) cm.
• Option 4: \(v_4\approx3.33\) (speed before \(40 - 50\) cm interval), \(v_5 = 5\) (speed in \(40 - 50\) cm interval). As \(v_5>v_4\), the car was not decelerating from \(40 - 50\) cm.

Answer:

The car was accelerating from \(20 - 30\) cm.