High School

A car accelerated from 88 feet per second (fps) to 220 fps in exactly 3 seconds. Assuming the acceleration was constant, what was the car's acceleration, in feet per second per second, from 88 fps to 220 fps?

F. [tex]\frac{1}{44}[/tex]
G. [tex]29 \frac{1}{3}[/tex]
H. 44
J. [tex]75 \frac{1}{3}[/tex]
K. [tex]102 \frac{2}{3}[/tex]

Answer :

To find the car's acceleration from 88 feet per second (fps) to 220 fps over a time period of 3 seconds, we need to use the formula for acceleration. The formula is:

[tex]\[ \text{acceleration} = \frac{\text{final velocity} - \text{initial velocity}}{\text{time}} \][/tex]

1. Identify the initial conditions:
- Initial velocity ([tex]\(v_i\)[/tex]) = 88 fps
- Final velocity ([tex]\(v_f\)[/tex]) = 220 fps
- Time ([tex]\(t\)[/tex]) = 3 seconds

2. Substitute these values into the formula:

[tex]\[ \text{acceleration} = \frac{220 \, \text{fps} - 88 \, \text{fps}}{3 \, \text{seconds}} \][/tex]

3. Perform the subtraction in the numerator:

[tex]\[ 220 \, \text{fps} - 88 \, \text{fps} = 132 \, \text{fps} \][/tex]

4. Divide the result by the time:

[tex]\[ \text{acceleration} = \frac{132 \, \text{fps}}{3 \, \text{seconds}} = 44 \, \text{fps}^2 \][/tex]

So, the car's acceleration is 44 feet per second squared, which corresponds to option H.

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