Question

A 3-column table with 1 row. The first column titled distance travelled (meters) has entry 6. 1. The second column labeled lower track elapsed time (seconds) has entry 4. 92. The third column labeled higher track elapsed time (seconds) has entry 3. 36. 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

Answer:

Based on the data provided by the entries in the columns of the table, the car was found to be faster on the upper track than on the lower track.

There are two possible reasons for this:

1)The upper way is wider, so the circulation is more fluid.

2)The upper track is straight and therefore shorter than the lower track.

• Rail car speed on the track:

Formula:

S=d/t

S=speed

D=distance traveled

T=elapsed

6.1/4.92=1.24m/s=4.46km/h

• Car speed on the top track

6.1/3.6=1.82=6.55km/h

The car is faster on the track of the upper track compared to that of the lower track of:

6.55-4.46=2.09km/h

Explanation:

To calculate a speed in meters/second, the distance in meters is divided by the time in seconds. If you want to convert a speed in m/s to km/h, just multiply the speed in m/s by 3.6 because 1 hour = 3600 seconds and 1 kilometer = 1000 meters.

Answer 2

Answer:

1) car travel faster on the higher track

2) shorter

3) 0.56m/s

Explanation:

got it right