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Basic Kinematics Questions, first year physics

  1. Sep 8, 2012 #1
    An object's motion can be described by a function that relates its position x (in meters) to time t (in seconds).

    Let this function be x = 6t2 + 2t + 4

    What is this object's acceleration?

    I tried inputting 6, but it counted it wrong.
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    "Stopping distance" is defined as the distance a car travels between the instant a driver sees a need for emergency stop and the instant that a car comes to rest when the brakes are fully applied. When an obstacle is first seen by a driver, the driver takes a certain time to react (traveling the "thinking distance") before applying the brakes. Then, the car must decelerate to a stop while the brakes are fully applied (traveling the "braking distance").

    Consider a driver traveling with a speed of 40 m/s. A child runs out into the roadway in front of the driver, and the driver takes 0.7 seconds to react to this need to suddenly stop. The brakes provide a constant deceleration rate of -10 m/s2. How far does the car travel before coming completely to rest?

    A full explanation on this question would be extremely helpful, thanks.

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    A car traveling to the right with initial speed 28 m/s must come completely to rest over a distance of 49 meters in order to pass a brake test. What acceleration must the car have? (Let the rightward direction be considered positive.)

    I'm confused on what equation I should use to answer this one/

    Equations:
    v= v(naught) + at

    x = x(naught) + v(naught)t + (1/2)at^2

    v^2 = v(naught)^2 + 2a (x-x(naught))

    Thank You,
     
  2. jcsd
  3. Sep 8, 2012 #2
    Do u know application of derivatives??
    if yes then just differentiate the equation with respect to time 2 times to get acceleration

    next one

    using v2-u2=2(a)(s)

    initial velocity is given
    final velocity is 0
    acceleration is given(-ve in this case)
    so find s or distance traveled after breaks were applied

    now u add the distance it traveled in 0.7 sec (before breaks were applied)i.e. initial velocity*time

    now add both displacements to get answer

    last question i will let u solve
     
  4. Sep 8, 2012 #3
    How you get the value 6?
    -----------------------
    Draw a velocity vs. time
    -------------------------
    As above. The area bounded by f(t), x-axis and y-axis is distance travelled.
     
  5. Sep 8, 2012 #4
    I'm currently in pre-calculus, so I haven't learned how to use derivatives, can you give me an alternative solution to the first question?

    For the second question, there are two acceleration givens, 40 and -10, which do I plug into the equation, and what does U stand for?

    Some sort of explanation on the third question would be extremely helpful too thanks.
     
  6. Sep 8, 2012 #5
    Hey second q there is only one acceleration that is -10m/s2
    40m/s is the value of u or initial velocity of car(v0)

    3rd question is just similar

    initial velocity is given (28 m/s)
    final velocity is 0
    distance to be traversed is given (49m)

    so use the equation
    v2-u2=2(a)(s)[u is initial velocity]

    u should get -ve acceleration
     
  7. Sep 9, 2012 #6
    I got 108 m/s^2 for the 2nd question, can you tell me if it's correct, I only have one more submission left to get the question correct.

    Here's my work:
    40^2=2(-10)s

    1600= -20s

    s1 = 80

    s2 = 0.7 x 40

    s2 = 28

    Answer = 108 Meters
     
  8. Sep 9, 2012 #7

    SammyS

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    Yes, 6 is wrong. By the way, when you do get the correct answer, it should include units.

    Do you know any kinematic equations that are valid for constant acceleration and include position and acceleration?
     
  9. Sep 9, 2012 #8
    Looks correct numerically but i think something is wrong

    If a car takes 108 meters to stop while moving then most probably the accident is inevitable (108 meters is not a small distance right??)

    What answer did u get on 3rd??
     
  10. Sep 9, 2012 #9
    -8 m/s^2 was my 3rd answer

    So is the 2nd q correct, regardless of whether or not the car hits the kid?
     
  11. Sep 9, 2012 #10
    Yes all are correct

    but did u get the first one??
     
  12. Sep 9, 2012 #11

    SammyS

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    Well, I see in your Original Post that you have:
    x = x0 + v0 t + (1/2)at2

    Compare this with
    x = 6t2 + 2t + 4​
    and pick out x0, v0, and a --- maybe it'sier to pick out (1/2)a .
     
  13. Sep 9, 2012 #12
    Thats a great way of solving man...

    that way we get a=12m/s2

    differentiating it 2 times with respect to time also gives us 12m/s2

    so i guess it is correct
     
  14. Sep 9, 2012 #13
    Can you guys helm me with this question? I was able to get the first part asking for average velocity, but this part is confusing me.

    http://www.webassign.net/userimages/Frensley_1D-Motion_Graph_001a.gif?db=v4net&id=124057

    The url has the link to the original velocity vs. time graph.

    Draw a graph of the car's position as a function of time for the first 15 seconds of its motion. In the table below, six individual times are selected. Fill in the table with the position of the car (in meters) at those times.

    Time (s) Position (m)
    0 12
    3

    6

    8

    12

    13

    15

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    Also, parabolic motion is kinda throwing me off when they use acceleration instead of velocity.

    A rocket fires from rest with an upward acceleration of 20 m/s2 for 2 seconds. After this time, the engine shuts off and the rocket freely falls back to the surface of the Earth.

    (b) Determine the height of the rocket when the engines shut off.


    (c) Determine the maximum height that the rocket reaches.



    (d) Determine the amount of time the rocket is in the air.


    Thanks,
     
  15. Sep 9, 2012 #14
    Hmm i think u need the initial speed with which rocket was projected....
    or was the body given acceleration from rest??
     
  16. Sep 9, 2012 #15
    I think it was given acceleration from rest, what would I need to do after that though?
     
  17. Sep 9, 2012 #16

    CAF123

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    The height of the rocket before the engine is switched off can be found using one of the kinematic relations. You know [itex] v_{oy} = 0, [/itex] so you can also find it's velocity after 2 seconds. After the engines go off, there is no longer any thrust and so only gravity acts on it. The rocket's maximum height will be when [itex] v = 0 [/itex] so using the velocity it had right before the engines were turned off, you can deduce this 'extra' height. Add this 'extra' height to the height gained in the acceleration phase to find the max height.
     
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