1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

In integral calculus

  1. Jun 21, 2004 #1
    1. What constant negative acceleration will enable a driver to decrease the speed from 120 km/hr to 60 km/hr while traveling a distance of 100m?

    Where I got stuck [WIGS]: Let us say you don't know the formula used in kinematics... the acceleration is unknown. Let acceleration be a... so,
    v = at + C(1), where C(1) is the first constant of integration and
    s = 0.5at^2 + C(1)t + C(2) where C(2) is the second constant of integration.

    Where will I evaluate my given values of velocity and distance? How can I
    get the value of the acceleration?

    2. A space shuttle climbs vertically with a constant acceleration of 10 yd/sec^2. If a radar-tracking dish, 1200 yd. from the shuttle's launch pad, follows the shuttle, how fast is the radar dish revolving 8 sec. after blast off?

    [WIGS]: from the manual, the answer is 0.06 rad/sec. But what I don't
    unerstand is how to arrive at that answer. To analyze the problem, it looks like a right triangle, your base is 1200 yd... how will antidifferentiation take place here? =)
  2. jcsd
  3. Jun 21, 2004 #2
    The trick here is that you can evaluate the constants of integration by letting t equal zero.

    So, starting from the definition of acceleration:

    [tex]a = dv/dt[/tex]

    [tex]a~dt = dv[/tex]

    [tex]\int a~dt = \int dv[/tex]

    [tex]at + C_1 = v[/tex]

    If t = 0, then [itex]C_1 = v[/itex]. The value of v when t = 0 must be [itex]v_0[/tex] by definition (the "initial" velocity), so:

    [tex]at + v_0 = v[/tex]

    Using the definition of velocity and integrating again:

    [tex]at + v_0 = dx/dt[/tex]

    [tex]at~dt + v_0~dt = dx[/tex]

    [tex]\int at~dt + \int v_0~dt = \int dx[/tex]

    [tex]\frac {at^2}{2} + v_0 t + C_2 = x[/tex]

    We use the same trick as before - at t = 0, [itex]C_2 = x[/itex], and we know that, by definition, [itex]x = x_0[/itex] at t = 0 (the initial position), so [itex]C_2 = x_0[/itex] and:

    [tex]\frac {at^2}{2} + v_0 t = x - x_0[/tex]

    To eliminate t, we take the equation we got after the first integration, solve for t, substitute into the most recent equation, simplify, and rearrange the result:

    [tex]v = v_0 + at[/tex]

    [tex]\frac {v - v_0}{a} = t[/tex]

    [tex]\frac {a}{2} \cdot \frac{v^2 - 2vv_0 + v_0^2}{a^2} + \frac {vv_0 - v_0^2}{a} = x - x_0[/tex]

    Multiply through by 2a and simplify:

    [tex]v^2 - v_0^2 = 2a(x - x_0)[/tex]

    [tex]\frac {v^2 - v_0^2}{2(x - x_0)} = a[/tex]

    Hopefully you don't have to prove your kinematics equations on every problem :wink:

    I haven't tried the second question yet, but it sounds like you can integrate the acceleration of the shuttle to find out how quickly that side of the triangle is growing with respect to time, then treat it as a related rates problem.
  4. Jun 21, 2004 #3
    1) You know that v(0) = 120 km/h, and s(0) = 0.

    2) Since you know the acceleration, you can calculate (with integration) how far the space ship has flown at the time t (call this s(t)). Let alpha be the angle between the radar dish and the shuttle, then tan(alpha) = s(t)/1200. Then you can calculate how fast alpha changes at t = 8, using the derivative...

    *edit* Oops, I see someone's already answered. That's what I get for taking so long when typing up my answer :P
  5. Jun 21, 2004 #4
  6. Jun 21, 2004 #5
    Yes, at t = 0, the change in displacement (x - x0) is zero - the driver hasn't had a chance to move yet, since no time has passed. x is the current displacement, and x0 is the initial displacement. At any other time, x - x0 is not zero.

    The problem says that the driver has travelled through a distance of 100m. We don't immediately know how long it took him to do that, but it doesn't matter - we know x0 = 0, x = 100 m, and so x - x0 = 100 m.
  7. Jun 22, 2004 #6
    Ah, I see. =)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: In integral calculus
  1. Calculus 2 Integral (Replies: 5)

  2. Integral calculus (Replies: 3)