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Calculate the average travel time for each distance, and then use the results to calculate.

The average time that it takes for the car to travel the first 0.25m is ___

s.


The average time to travel just between 0.25 m and 0.50 m is ____

s.


Given the time taken to travel the second 0.25 m section, the velocity would be ___

m/s.

Calculate the average travel time for each distance, and then use the results to calculate-example-1
User GrayFox
by
5.7k points

2 Answers

6 votes

Answer:

The average time that it takes for the car to travel the first 0.25m is

2.23 s.

The average time to travel just between 0.25 m and 0.50 m is

0.90 s.

Given the time taken to travel the second 0.25 m section, the velocity would be

0.28 m/s.

Step-by-step explanation:

Just did this problem! :)

User Sbonkosky
by
6.8k points
1 vote

The average time that it takes for the car to travel the first 0.25m is 2.23 s

The average time that it takes for the car to travel the first 0.25 m is given by:


t=(2.24 s+2.21 s+2.23 s)/(3)=2.227 s \sim 2.23 s

The average time to travel just between 0.25 m and 0.50 m is 0.90 s

First of all, we need to calculate the time the car takes in each trial to travel between 0.25 m and 0.50 m:


t_1 = 3.16 s - 2.24 s=0.92 s\\t_2 = 3.08 s- 2.21 s=0.87 s\\t_3 =3.15 s- 2.23 s=0.92 s

Then, the average time can be calculated as


t=(0.92 s+0.87 s+0.92 s)/(3)=0.90 s

Given the time taken to travel the second 0.25 m section, the velocity would be 0.28 m/s

The velocity of the car while travelling the second 0.25 m section is equal to the distance covered (0.25 m) divided by the average time (0.90 s):


v=(d)/(t)=(0.25 m)/(0.90 s)=0.28 m/s

User Aaron Breckenridge
by
5.3k points