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Rearranging equations of accelerated motion

  1. Oct 2, 2011 #1
    1. The problem statement, all variables and given/known data
    theres two questions actually.

    1) a boat increases its speed from 5.0 m/s to 7.5 m/s over a distance of 50.0m. what it the boat's acceleration?

    2) within 4.0s of liftoff, a spacecraft that is uniformly accelerating straight upward from rest reaches an altitude of 4.50 x 10^2 m [up]
    a) what is the spacecraft's acceleration
    b)at what velocity is the spacecraft travelling when it reaches this altitude?


    2. Relevant equations

    for 1) i just need to rearrange vf^2 = vi^2 +2aΔd to find acceleration

    for 2a) the same thing but with the equation, Δd=viΔt + 1/2aΔt^2



    3. The attempt at a solution

    ive tried rearranging the equation but i get stuck and i dont know what to do next. please help
     
  2. jcsd
  3. Oct 2, 2011 #2

    You want to use algebra to isolate the variable a, that is, to get it all alone on one side of the equation. Since it is an equation you can perform the same operation to both sides of the equal sign to change its form but not its meaning. For example, if you were trying to isolate z in [itex] x = p + dz [/itex] you would start by subtracting p from both sides. This leaves you with [itex] x - p = dz [/itex]. Now to isolate z we can divide both sides by d to get [itex] \frac{x - p}{d} = z [/itex]. Can you apply a similar process to [itex] v^2 = v_0^2 + 2 a \Delta x [/itex] to isolate a?
     
  4. Oct 2, 2011 #3
    Im not really good at this but i tried:

    vf^2 = vi^2 + 2a Δd

    vf^2/2a = vi^2 + Δd

    am i doing this right? I just want to know how to get rid of vf^2/2a so i can isolate 2a.
     
  5. Oct 2, 2011 #4
    That's not quite right. Remember that if you want to divide through by "a" you need to do it to every term. What if you started by subtracting vi^2 from each side of the equation and then diveded through by 2Δd? Dividing by "a" first is not that helpful.

    I'll show you the first step,

    [tex] vf^2 = vi^2 + 2a Δd [/tex]
    subtract [itex] vi^2 [/itex] from both sides
    [tex] vf^2 - vi^2 = vi^2 - vi^2 + 2a \Delta d [/tex]
    [tex] vf^2 - vi^2 = 0+ 2a \Delta d [/tex]
    [tex] vf^2 - vi^2 = 2a \Delta d [/tex]

    Now to get a all by itself we need to divide both sides by 2Δd. Can you try this and repost if you are still not sure?
     
  6. Oct 2, 2011 #5
    no i got it :D thanks a lot . I was probably just thinking too much but now i have another problem. If you dont mind can you help me with that question?

    Two go carts A and B race each other around a 1.0 km track. Go-cart A travels at a constant speed of 20.0 m/s. Go cart B accelerates uniformly from rest at a rate of 0.333 m/s^2. Which go-cart wins the race and by how much time?

    The answer at the back says A wins by 27 s.

    For this question, I know that Δd = 1000 m
    a= 0.333 m/s^2
    and that were looking for Δt.
    I just dont know how to go about. I was thinking that I have to use two separate equations.
     
  7. Oct 2, 2011 #6
    You're right in that a good approach is to find the time it takes each car to travel the track and compare them (Δt for each). You can use this equation

    [tex]
    x = x_0 + v_0 t + (1/2) a t^2
    [/tex]

    In the case of no acceleration "a" would just be zero.
     
  8. Oct 2, 2011 #7
    give me another hint? i dont get it :s
     
  9. Oct 2, 2011 #8

    PeterO

    User Avatar
    Homework Helper

    If you can't re-arrange the formula with all the terms in it, substitute the known values then just solve the problem with only 1 pro-numeral [a] and a whole lot of digits.

    Once the numbers are substituted, this becomes a Junior High school equation to solve.
     
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