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Centripetal and Tangential Acceleration

  1. Feb 3, 2016 #1
    1. The problem statement, all variables and given/known data
    A car drives on a circular road with radius ##R##. The distance driven by the car is given by ##d(t) = at^3 + bt## [where ##t## in seconds will give ##d## in meters]. In terms of ##a##, ##b##, and ##R##, and when ##t = 2## seconds, find an expression for the magnitudes of (i) the tangential acceleration ##a_{tan}##, and (ii) the radial acceleration ##a_{rad}##.

    2. Relevant equations
    $$d(t) = at^3 + bt$$
    $$a_{rad} = \frac{v^2}{r}$$
    $$a_{tan} = \frac{d|\vec v|}{dt}$$

    3. The attempt at a solution
    For i) I took the second derivative of ##d(t)## and plugged in ##t = 2## to get:

    $$a_{tan} = 6a(2)$$

    My reasoning for this is that ##d(t)## describes the distance the car traveled around circular road, which means that its derivative will describe the car's tangential velocity at time ##t##, and ultimately means that its second derivative will be the tangential acceleration.

    For ii) I took the first derivative of ##d(t)## because, like with part i), I reasoned it must be the tangential velocity of the car since ##d(t)## describes the distance driven by the car along the curved road. So I first took the derivative of ##d(t)## and got:

    $$v(t) = 3a(2)^2 + b$$

    Since the formula for ##a_{rad}## is quotient of the square of the tangential velocity and the radius of the circle, I plugged ##v(t)## into ##v## and ##R## into ##r## for the formula of ##a_{rad}## to get:

    $$a_{rad} = \frac{(3a(2)^2 + b)^2}{R}$$

    The issue I'm having with this problem is that I don't meet the conditions stated in the problem for part i): I don't have ##a_{tan}## in terms of ##a##,##b##, and ##R##. So what I'm left wondering is whether ##v(t)## actually is the tangential velocity or if I need to use another formula that will help me meet the conditions stated in the problem. Since ##a(t)## is not zero I'm assuming the car isn't moving at uniform circular motion, so uniform circular motion equations don't apply.
     
  2. jcsd
  3. Feb 3, 2016 #2

    ehild

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    The time derivative of the distance traveled is speed, a scalar, not 'tangential velocity' which is a vector. But you need speed to get the radial acceleration. The time derivative of speed is the tangential acceleration. So your derivation is correct. The problem asked the tangential and radial acceleration in terms of a, b and R. You did it. If a variable does not appear in a formula, you can consider that it is with zero coefficient somewhere. If d(t) was (at^3+bt)^2, for example, both a and b would appear in the tangential acceleration.
     
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