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Boat acceleration physics problem

  1. May 22, 2004 #1
    Hi, I have a physics test on Monday and would like some help with these problems.

    Kinematics problems:

    1. A boat starting from rest on a lake accelerates in a straight line at a constant rate of 2.0 m/s^2 for 6.0 s. How far does the boat travel during this time?

    2. A bullet traveling horizontally at a speed of 350 m/s hits a board perpendicular to the surface, passes 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?

    Work and Energy problem: (equations: W=Fdcos(theta), Wnet = change in K, U= mgh, K=1/2mv^2, E = K + U, Fnet=ma, Fg = mg)

    1. A 500 kg helicopter ascends from the ground with an acceleration of 2.00 m/s^2. Over a 5.00-s interval, what is the work done by the lifting force?

    2. An automobile with a mass of 1200 kg travels at a speed of 90 km/h. What force would be required if it took 100 m to stop?

    3. A person standing on a bridge at a height of 80 m above a river drops a 0.250-kg rock. (a) What are the rock's kinetic and potential energy at the time of release relative to the surface of the river? (b) What are the energies just before the rock hits the water?
  2. jcsd
  3. May 23, 2004 #2


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    The 2*(1.+2.) problems should all be solved under the assumption of constant acceleration.
    Clarify to yourself what impact this has for the solution in each particular case.

    Problem 3. should be solved using conservation of mechanical energy.
    In particular, initial kinetic energy is zero (the rock is "dropped"), while final potential energy is zero (you are given that potential energy is measured relative to the surface of the river)
  4. May 23, 2004 #3
    Q1: You need to use, and get well aquainted with the equations of motion. Its one of the easyer things to learn. Basicly free marks on an examination if you remember the rules and understand what is happening.

    the equations are (forgive me for writing them as they are but i dont know how these other members set it out neatly)

    v = u + at
    v^2 - u^2 = 2as
    s = (1/2)(u+v)t
    s = ut + (1/2)at^2
    s = vt - (1/2)at^2

    u = initial velocity
    v = final velocity
    a = accelaration (constant)
    t = time taken
    s = distance traveled.

    so, your boat starts from rest. The initial velocity is 0m/s.
    your boat has a constant accelaration of 2m/s^2
    the time is takes for this motion is 5s
    and you want to find the distance.

    the 4 lines above correspond to 4 variables. 3 of which you have. Pick one of the equations that contain the 4 variables and solv to find the unknown. Then remember the equations!

    use the equations for Q2 as well. There are again 4 variables, just find an equation they fit into.

    as for the work/energy problems. Using Energy is sometimes much more convientiant than using force's. So try and remember how to use Energy to solve problems.

    Work is equal to force * displacement
    so, you have a mass(M) of 500kg accelarating(a) up at 2m/s^2
    Use newtons second law of motion to find the force F (F=Ma)
    the helicopter starts from the ground, and moves up with accelaration of 2m/s^2 over a time of 5s. So use a equation of motion to find the distance traveled (displacement)
    finaly find the work from W=Fd

    For Q2, it can be done with force fairly easily. Use F=ma and an equation of motion

    For Q3, potential energy is equal to mgh (mass * acceleration due to gravity * height)
    Kinetic enrgy is equal to (1/2)mv^2 (1/2 * mass * velocity^2)

    so, if the rock is not moving, its velocity is 0. What does this make the velocity when he is holding it on the bridge?
    But when its just about to hit the water, its moving quickly..so there must be some kinetic energy.

    opposite to this, just before it hits the water, the rock is 0 meters (height) from the water. So what would be the potential energy just before it hit the water.
    When he is holding the rock on the bridge, its height is a maximum. So the potential energy must be at its maximum.

    If you have a look at the behaviour of the two energy's you will notice as one decreases, the other decreases. This is conservation of energy and is a very important part of physics. No energy can be created or deleted, it can only be changed.
    So, all the potential energy and kinetic energy of the rock when the boy is holding it MUST BE converted to potential and kinetic energy just before the rock hits the water.

    you can write this as
    (energy top = energy bottom)
    mgh + (1/2)mv^2 = mgh + (1/2)mv^2
    and noting that the kinetic energy at the top is 0 and the potential energy at the bottom is 0
    mgh (top)= (1/2)mv^2 (bottom)

    thats some physics lesson for you. Cant explain it the best over the internet. Hope it helps.
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