Asked by Gabrielle
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.
If Mason's car takes 35.9 meters to stop, what is the acceleration of the car?
--
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.
Answers
Answered by
bobpursley
No,you get a that is negative, and you didn't say that in your answer.
Answered by
Gabrielle
Oh, wow. Talk about a dumb mistake. Thank you bobpursley!!
Answered by
bobpursley
thanks for showing your work. It is easy to critique work, most kids rely on my crystal ball, which does not always work.
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