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anonymous

  • 5 years ago

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 40 m in front of you. Your reaction time before stepping on the brakes is 0.50 s , and the maximum deceleration of your car is 12 m/2^2 How much distance is between you and the deer when you come to a stop? What is the maximum speed you could have and still not hit the deer?

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  1. anonymous
    • 5 years ago
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    I solved the problem, but I am not so sure about my answer.

  2. anonymous
    • 5 years ago
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    cam i see wat you did?

  3. anonymous
    • 5 years ago
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    yeah sure!

  4. anonymous
    • 5 years ago
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    I first found the time needed for the car to stop, that's when the final velocity is zero. Apply it to the formula: \[v_f=v_i+at \implies 0=20-12t \implies t=20/12=1.67\] (since it's decelerating, I took a to be negative)

  5. anonymous
    • 5 years ago
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    but we know that the reaction was 0.5 s late so the total time needed is t=1.67+0.5=2.17

  6. anonymous
    • 5 years ago
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    are you following so far?

  7. anonymous
    • 5 years ago
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    yes so far so good

  8. anonymous
    • 5 years ago
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    ok I will just write the rest. now we will have to find the distance he is going to drive during this period of time, which is: \[x={v_i}t+0.5at^2\] just direct substitution in the formula, you will get: \[x=20 (2.17)+0.5(-12)(2.17)^2=15.15 m\] so the distance he traveles before the car stops is 15.15 meter

  9. anonymous
    • 5 years ago
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    now it's easy to find the distance between the car and the deer, which is 40-15.15=24.85 meter

  10. anonymous
    • 5 years ago
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    I dunno if you have the final answers?

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