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?

Answer 1

I don't see the table or the numbers you mentioned. I can answer conceptually and show exactly how to compute the speeds and the percent difference — if you paste the times (and the track lengths, if they are different) I will compute the numeric answers for you.

Conceptual answer
- If the elapsed time on the higher track is smaller than on the lower track (for the same distance traveled), the car is faster on the higher track. Shorter time for the same distance means larger average speed.

How to compute it
1. For each track, compute average speed = distance / time.
- v_low = d_low / t_low
- v_high = d_high / t_high
(If both tracks have the same length, use the same d for both.)
2. Absolute difference: Δv = v_high − v_low.
3. Relative (percent) faster: percent faster = (Δv / v_low) × 100% (or (v_high / v_low − 1) × 100%).

Example (illustration)
- Suppose both tracks are 2.0 m long, t_low = 4.0 s, t_high = 3.0 s.
- v_low = 2.0 / 4.0 = 0.50 m/s
- v_high = 2.0 / 3.0 ≈ 0.667 m/s
- Δv = 0.667 − 0.50 = 0.167 m/s
- percent faster = (0.167 / 0.50) × 100% = 33.4%

If you paste the table (track lengths and elapsed times), I will compute the exact speeds and how much faster the car was on the higher track.