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Kinematics (constant acceleration)

  1. Sep 21, 2006 #1
    I have three problems that have stumped me. I attempted to utilize the equations my teacher said we'd be using but I don't know where I went wrong or what each equation is specifically for (e.g. finding displacement in constant acceleration or whatnot). Am I using the equations correctly? :uhh:

    1st Problem: A car accelerates from rest at a constant rate of 2.0 m/s^2 for 5.0s. A) What is the speed of the car at that time? B) How far does the car travel in this time?

    My workings:

    x = Vot + 1/2at^2
    x = 0(5.0s) + 1/2(2.0 m/s^2)(5.0s)^2
    = 0 + 1/2(2.0 m/s^2)(25.0 s^2)
    = 50 m/s^4/2
    = 25 m/s^4

    2nd Problem: A bullet traveling horizontally at a speed of 350 m/s hits a board perpendicular to the surface, passed through it, and emerges on the other side at a speed of 210 m/s. If the board is 4.00 cm thick, how long does the bullet take to pass through it?

    My workings:

    I know that Vo or Vi is 350 m/s, A is 4.00 cm, Vf is 210 m/s but how do I convert the negative acceleration factor, A, into a speed? I also have no clue as to which equation to use for this.

    3rd Problem: Two identical cars capable of accelerating at 3.00 m/s^2 are racing on a straight track and cross the starting line together with running starts. Car A has an initial speed of 2.50 m/s; car B starts with an initial speed of 5.00 m/s. A) What is the separation of the two cars after 10 m/s. B) Which car is moving faster after 10 s?

    My workings:

    By the separation of the two cars, do they mean displacement? I don't know where to proceed. So far..

    A - 3.00 m/s^2 (I'm not sure of this as the problem used the word "capable")
    Vo (A) - 2.5 m/s
    Vo (B) - 5.00 m/s
    T - 10 s
    Vf - ?

    When I use x=VoT + 1/2at^2 my result ends up with m/s^4. Am I using the wrong equation?
     
  2. jcsd
  3. Sep 21, 2006 #2
    For the first problem you got worng units, i.e., are you sure that m.s-2.s2 is equal to m.s-4? Nevertheless, the answer is correct.
    Also you did not solve (A). For that purpose the equation v(t) = v0 + at might be very useful.

    For the second problem: do you know the equation vf2 = v02 + 2ax or how to work it out?
     
  4. Sep 21, 2006 #3

    Andrew Mason

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    Distance is in metres, not metres/s^4. So the correct answer to B) is 25 m. I assume you had no difficulty with A).

    It always helps to draw a graph of speed vs. time. The area under the graph over a time interval is the distance covered in that time interval. Assume a constant force (constant deceleration). Write the expression for d in terms of vi and vf and [itex]\Delta t[/itex] (just divide into a rectangle and a triangle). Solve for [itex]\Delta t[/itex]. You will see that this is the same as finding the average speed (since we are assuming constant deceleration, this is simply the average of the initial and final speeds) and dividing the distance by that average speed.

    Now see if you can work out the third question on your own.

    AM
     
    Last edited: Sep 21, 2006
  5. Sep 21, 2006 #4

    HallsofIvy

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    What does this problem ask you to find? Are you even aware that there are two questions- since you give only one answer?
    You say you are not sure what the equations are used for. I recommend that you explicitely write out what your variables represent- and include the units.
    What is "x" in this problem? What is t? You clearly know that a is the acceleration, in m/s2 since you substitute the given 2.0 m/s2 and apparently know that t is time, in seconds, since you use that. But is x what you want to find? Oh, and while it is a very good thing to "calculate" with the units, be sure that you do it correctly: [itex]\frac{m}{s^2}(sec^2)[/itex] is not equal to [itex]m/sec^4[/itex]! In fact, you should know that there is no quantity measured in [itex]m/sec^4[/itex]. In fact, the "s2"s cancel and [itex]\frac{m}{s^2}(sec^2}[/itex] gives m. That's correct because the x in this equation is the distance, in meters, that the object has moved. But (A) asks for speed, not distance. It is (B) that asks for the distance.

     
  6. Sep 21, 2006 #5
    There was no need to be so aggressive and harsh with me.

    Also, I tried to use A, 4.00 cm, as a factor for deceleration, thus my confusion.
     
  7. Sep 21, 2006 #6

    Andrew Mason

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    I am quite sure that Halls was not trying to be harsh or aggressive. I think if you reread his posts you will see that he is pointing out some very basic errors you are making with units.

    When you speak about distance the units must be in distance (eg. m) not time. When you speak about speed, units must be distance/time (eg. m/sec). For acceleration, units must be in speed/time = distance/time^2. No one here will understand what you mean when you say A = 4.00 cm if A is acceleration, which is what you seem to be saying.

    In problem 2, the speed change is 140 m/sec. The average speed
    through the board is __________? Average speed is total distance over total time, so if you work out the average speed, you can determine the time. The acceleration is change in speed divided by the time interval, so you can work that out too. Just make sure you have the units right.

    AM
     
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