Results 1 to 7 of 7

Math Help - rate

  1. #1
    Newbie
    Joined
    Jun 2007
    Posts
    18

    rate

    1. (a) A car traveling at a speed of v can brake to an emergency stop in a distance x. Assuming all other driving conditions arte all similar, if traveling speed of the car doubles, the stopping distance will be (1) 2x, or (2) 2x, or (3) 4x: (b) a driver traveling at 40.0 km/h in a school zone can brake to an emergency stop in 3.00 m. What would be the braking distance if the car were traveling at 60.0 km/h?

    2. A student drops a ball from the top of a building; it takes 2.8 s for the ball to reach the ground. (a) What is the ball's speed just before hitting the ground? (b) What is the height of the building?

    3. A photographer in a helicopter ascending vertically at a constant rate of 12.5 m/s accidentally drops a camera out the window when the helicopter is 60 m above the ground. (a) How long will it take the camera to reach the ground? (b) What will it's speed be when it hits?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member Jonboy's Avatar
    Joined
    May 2007
    From
    South Point
    Posts
    193
    I'm sure this requires nothing more than some formulas.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,707
    Thanks
    627
    Hello, John!

    For the last two, we are expected to know the "free fall" formula: . y \; = \; h_o + v_ot - gt^2

    . . where: . \begin{array}{ccc}h_o & = & \text{initial height} \\ v_o & = &\text{initial velocity} \\ g & = & \text{gravitational constant} \\ y & = & \text{height of object} \end{array}

    Note: g is usually 16 ft/secē or 4.9 m/secē.



    2. A student drops a ball from the top of a building;
    it takes 2.8 s for the ball to reach the ground.
    (a) What is the ball's speed just before hitting the ground?
    (b) What is the height of the building?
    We have: . v_o = 0 . (the ball is dropped, not thrown).

    The equation is: . y \;=\;h_o - 16t^2

    (a) The velocity is given by the derivative: . v(t) \:=\:y' \:=\:-32t

    Then: . v(2.8) \:=\:-32(2.8) \:=\:-89.6

    The ball's speed on impact is 89.6 feet per second (downward, of course).


    (b) When the ball strikes the ground, its height is zero.
    . . We have: . 0 \;=\;h_o - 16\cdot2.8^2\quad\Rightarrow\quad h_o \:=\:125.44

    The height of the building is 125.44 feet.




    3. A photographer in a helicopter ascending vertically at a constant rate of 12.5 m/s
    accidentally drops a camera out the window when the helicopter is 60 m above the ground.
    (a) How long will it take the camera to reach the ground?
    (b) What will its speed be when it hits?
    We are given: . h_o = 60,\;v_o = 12.5

    The equation is: . y \;=\;60 + 12.5t - 4.9t^2

    (a) "reach the ground" means: y = 0
    We have: . 60 + 12.5t - 4.9t^2 \:=\:0
    . . which factors: . (t - 5)(4.9t + 12) \:=\:0
    . . and has the positive root: . t = 5
    Therefore, it takes the camera 5 seconds to crash to the ground.

    (b) The velocity is: . v(t) \;=\;y' \;=\;12.5 - 9.8t
    . . Hence: . v(5)\;=\;12.5 - 9.8(5) \;=\;-36.5
    Therefore, its speed on impact is 36.5 meters per second.

    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jul 2007
    Posts
    90
    aaaand...

    Problem 1,a
    We are told that the square of velocity is proportional to the stopping distance. (look at topsquark's post for its justification)
    v^2 \propto x so (2v)^2=4v \propto 4x.

    4x

    Problem 1,b
    40^2=1600\frac{km}{h} \propto 3.00m

    Using this information, we can find the stopping distance of the same car at 60.0 km/h by using a proportion and solving for x.

    \frac{3}{1600} = \frac{x}{3600} \Longleftrightarrow x=6.75m


    These veterans are just too fast...
    Last edited by rualin; July 5th 2007 at 04:45 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member Jonboy's Avatar
    Joined
    May 2007
    From
    South Point
    Posts
    193
    Quote Originally Posted by rualin View Post
    These veterans are just too fast...
    Soroban is not the fastest so you haven't seen anything yet lol j/m. Soroban is one of the most cunning and interesting.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,889
    Thanks
    326
    Awards
    1
    Quote Originally Posted by rualin View Post
    Problem 1,a
    We are told that velocity is directly proportional to the stopping distance.
    No we weren't. Try this programme:
    Find the acceleration for when it starts at speed v.
    The acceleration will be the same as this for when it starts at 2v.

    In both cases, apply the formula:
    v^2 = v_0^2 + 2a(x - x_0)

    -Dan
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Jul 2007
    Posts
    90
    Thanks, topsquark! I realize my mistake and have corrected it... I hope.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Communication channel: baud rate, bit rate, etc.
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: July 3rd 2011, 08:16 AM
  2. Replies: 3
    Last Post: April 12th 2011, 09:51 AM
  3. Replies: 2
    Last Post: October 15th 2009, 06:06 AM
  4. Average rate and instantaneous rate?
    Posted in the Calculus Forum
    Replies: 10
    Last Post: August 30th 2009, 03:23 PM
  5. Rate?
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: May 24th 2009, 11:25 PM

Search Tags


/mathhelpforum @mathhelpforum