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Uniform acceleration problems

  1. Sep 24, 2007 #1
    I'm having trouble with two problems:

    1. A driver is travelling at 25 m/s when she spots a sign that reads "BRIDGE OUT AHEAD". It takes her 1.0 s to react and begin braking. The car slows down at a rate of 3.0 m/s^2.
    Luckily, she stops 5.0 m short of the washed out bridge.

    a) How much time was required to stop the car once the brakes were applied?

    My attempt:

    t= 0 m/s - 25 m/s (divided by) 3.0 m/s^2
    t= -8.3
    t =(8.3 s)


    b) How far was the driver from the bridge when she first noticed the sign?

    My attempt:

    d = 0.5(V1 + V2)t
    d = 0.5(0 + 25) 8.3
    d = 104 + 5.0 (she fell short)
    d = 109 m


    2. A box accidently falls from the back of a truck and hits the ground with a speed of
    15 m/s. it slides along the ground for a distance of 45 m before coming to rest.

    Determine:

    a) the length of time the box slides before stopping

    My attempt:

    d = 0.5(V1 + V2)t
    45 = 0.5(15 + 0) t
    solve for t and get
    t = 6


    c) the time it takes to slide the last 10 m

    I don't know this one.
     
  2. jcsd
  3. Sep 24, 2007 #2
    Can someone just quickly check if Im doing them right.
     
  4. Sep 24, 2007 #3

    bel

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    2c)Figure out the acceleration with the information you have and apply [tex]s=vt-\frac{1}{2}at^2[/tex] with s=10.
     
  5. Sep 24, 2007 #4
    Thanks :) I am I doing the other ones correctly?
     
  6. Sep 24, 2007 #5

    Doc Al

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    Staff: Mentor

    Don't forget that it takes her 1.0 second to react.
    Start by finding the acceleration.
     
  7. Sep 24, 2007 #6

    learningphysics

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    Homework Helper

    I'd do it like this (0-25)/(-3)... since it is slowing down acceleration is -3...

    You also have the 1.0s before she reacted... so you need to add 25m/s*1.0s = 25m

    Looks good.

    find the acceleration of the block...
     
  8. Sep 24, 2007 #7
    So I would subtract 1.0 s from 8.3?
     
  9. Sep 24, 2007 #8
    Thanks :) That helped a lot!
     
  10. Sep 24, 2007 #9

    Doc Al

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    Staff: Mentor

    No. The 8.3s is for the accelerated portion, which you've already figured out. How far does she go before she even hits the brakes?
     
  11. Sep 24, 2007 #10
    Doesn't say :S
     
  12. Sep 24, 2007 #11

    Doc Al

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    Staff: Mentor

    You know the speed and the time!
     
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