Question

based on the time measurements in the table, what can be said about the speed of the car on the lower track as compared to the higher track?
how can the reasoning for the above answer be best explained? on the higher track, the elapsed time is
calculate speeds for each track. how much faster was the car on the higher track than the lower track?

Explanation:

First Sub - Question: Speed Comparison

Step1: Recall Speed Formula

Speed is calculated as \( v=\frac{d}{t} \), where \( d \) is distance and \( t \) is time. For the same distance (\( d = 6.01\space m \)), a smaller time means a higher speed. The time on the lower track (\( t_{lower}=4.92\space s \)) is greater than the time on the higher track (\( t_{higher}=3.36\space s \)). So, the speed on the lower track is less than the speed on the higher track (or the car is slower on the lower track compared to the higher track).

Step2: Compare Speeds

Since \( v\propto\frac{1}{t} \) for constant \( d \), and \( t_{lower}>t_{higher} \), then \( v_{lower}

Step1: Analyze Time Values

We have \( t_{lower} = 4.92\space s \) and \( t_{higher}=3.36\space s \). Comparing these two values, \( 3.36<4.92 \), so on the higher track, the elapsed time is less than on the lower track.

Step2: Relate Time to Speed

Using the speed formula \( v = \frac{d}{t} \), for the same distance \( d \), when \( t \) is smaller, \( v \) is larger. So a smaller elapsed time on the higher track means a higher speed there.

Step1: Calculate Speed on Lower Track

Using \( v=\frac{d}{t} \), for the lower track, \( d = 6.01\space m \), \( t = 4.92\space s \). So \( v_{lower}=\frac{6.01}{4.92}\approx1.22\space m/s \).

Step2: Calculate Speed on Higher Track

For the higher track, \( d = 6.01\space m \), \( t = 3.36\space s \). So \( v_{higher}=\frac{6.01}{3.36}\approx1.79\space m/s \).

Step3: Calculate Speed Difference

Subtract the lower speed from the higher speed: \( \Delta v=v_{higher}-v_{lower}=1.79 - 1.22 = 0.57\space m/s \) (approximate values, more precise calculation: \( v_{lower}=\frac{6.01}{4.92}\approx1.2215\space m/s \), \( v_{higher}=\frac{6.01}{3.36}\approx1.7887\space m/s \), \( \Delta v = 1.7887 - 1.2215=0.5672\approx0.57\space m/s \))

Answer:

The car on the lower track is slower than the car on the higher track (or the speed of the car on the lower track is less than the speed on the higher track)

Second Sub - Question: Reasoning for the Answer