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Homework Help: Speed, Acceleration and Time - SUVAT Equations and Speed Time Graph

  1. Jun 11, 2012 #1
    1. The problem statement, all variables and given/known data

    In 4 seconds a car travels with constant velocity of 50m/s. After 4 seconds the car decelerates at 10m/s^2 until velocity = 0. What is the velocity at 8 and 12 seconds and what is the distance traveled after 12 seconds?

    Also how would I draw the acceleration time graph and relate that to a speed time graph?

    2. Relevant equations


    3. The attempt at a solution

    But this does not seem to give me the correct answer. I was told the total distance is 370 meters
  2. jcsd
  3. Jun 11, 2012 #2
    8 seconds total is 4 seconds after the deceleration starts. So the velocity at t=8s is equal to 50 m/s - 4 s ( 10 m/s2 ) = 10 m/s

    12-4=8 so 50 m/s - 8 s ( 10 m/s2 ) = -30 m/s, however the deceleration end at v= 0 m/s. So the speed actually is 0 m/s.

    To find the total distance traveled one has to find to the moment the car first has speed v=0. To do that solve the equation: 50 m/s - tstop ( 10 m/s2 ) = 0
    Solving gives tstop = 5 s (actually 5 s for the moment deceleration so 9 seconds total)
    The average speed traveled is (50 m/s + 0 m/s)/2 = 25 m/s (this work only if the deceleration or accelaration is constant)
    The distance traveled during deceleration is 25 m/s * 5 s = 125 meters
    The distance traveled before deceleration is 50 m/s * 4 s = 200 meters
    The total distance traveled is 325 meters.

    An acceleration vs time chart would look like this (x-axis should be t-axis):
    http://www4c.wolframalpha.com/Calculate/MSP/MSP20271a1ifb1cc14f2e4i000043e243867d21fe1a?MSPStoreType=image/gif&s=41&w=290&h=123&cdf=RangeControl [Broken]

    A speed vs time chart would look like this (x-axis should be t-axis):
    http://www4c.wolframalpha.com/Calculate/MSP/MSP731a1ifd56g88acbe20000156h79ga9hi13fed?MSPStoreType=image/gif&s=37&w=290&h=136&cdf=RangeControl [Broken]

    A location vs time chart would look like this:
    http://www4c.wolframalpha.com/Calculate/MSP/MSP6251a1ifd95dbc17f8b00006842i3gc0ig97h33?MSPStoreType=image/gif&s=26&w=287&h=127&cdf=RangeControl [Broken]
    Last edited by a moderator: May 6, 2017
  4. Jun 11, 2012 #3
    Zaal post is not correct.
    The question is the distance travelled not displacement.
    Try to draw the diagram.
    Start with constant velocity of 50m/s
    At t=4secs, draw line with gradient -10 till t=12sec.
    Add the areas(both sides taken as positive, since it is distance covered) and you get the answer.
  5. Jun 11, 2012 #4
    Sorry I made one mistake I corrected it (I forgot to add the distance traveled before deceleration).
    PS Don't forget v=0 for t>9
  6. Jun 11, 2012 #5
    Your graph should be extended till t=12sec
    The distance =4(50)+(50)(50)/20 +1/2(10)(3)2 =200+125+45=370m
  7. Jun 11, 2012 #6


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    Homework Helper

    You ignore the following sentence of the problem:

    The wording of the problem is confusing.

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