Question

I need help ASAP.

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

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

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

Answer 1

1. 2.23

2. 0.90

3. 0.28

Answer 2

1. Average time for the first 0.25 m: 2.23 s

Step-by-step explanation:

The average time that it takes for the car to travel the first 0.25 m is given by the average of the first three measures, so:

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

2. Average time to travel between 0.25 m and 0.50 m: 0.90 s

Step-by-step explanation:

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

t1 = 3.16 s - 2.24 s = 0.92 s

t2 = 3.08 s - 2.21 s = 0.87 s

t3 = 3.15 s - 2.23 s = 0.92 s

So, the average time is

`t=(0.92 s + 0.87 s + 0.92 s)/(3)=0.903 s=0.90 s`

3. Velocity in the second 0.25 m section: 0.28 m/s

Step-by-step explanation:

The average velocity in the second 0.25 m section is equal to the ratio between the distance covered (0.25 m) and the average time taken (0.90 s):

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