On Williamston Rd., Mason is driving 24.2 m/s. After spotting a large deer in the road ahead, Mason brakes the car to a complete stop.
If Mason's car takes 35.9 meters to stop, what is the acceleration of the car?
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Here's what I'm getting, but my program that the problem comes from says I'm wrong.
Using Vf^2=Vi^2+2ax, I put 0 for the final velocity, 24.2m/s for the initial velocity, and 35.9m for the distance(x). So I have 0=24.2^2+(2)(a)(35.9), which makes it 0=585.64+71.8a. Then, taking 585.64 from each side, I get -585.64=71.8a and get a=8.16m/s^2.
If you see something I'm doing wrong, please help me fix this.
No,you get a that is negative, and you didn't say that in your answer.
Oh, wow. Talk about a dumb mistake. Thank you bobpursley!!
thanks for showing your work. It is easy to critique work, most kids rely on my crystal ball, which does not always work.
Your calculations are correct, and the acceleration you obtained, a = 8.16 m/s^2, is also correct. There doesn't seem to be any mistake in your working.
It's possible that the program you're using might have a different answer or might require a specific format for the answer. It would be best to double-check the requirements of the program or consult the program's documentation to see if there are any specific conventions or rounding rules that need to be followed.
If you are confident in your calculations, you can try submitting your answer as is or round it to the required number of decimal places, if specified. If the program continues to mark your answer as incorrect, you may need to reach out to your instructor or the program's support team for further clarification.