? asked in Science & MathematicsPhysics · 1 decade ago

A electron moving along the x axis has a position given by x = 20 te-1.4 t m, where t is in seconds. How far i?

A electron moving along the x axis has a position given by x = 20 te-1.4 t m, where t is in seconds. How far is the electron from the origin when it momentarily stops?

Update:

X = 20 te^-1.4t

3 Answers

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  • 1 decade ago
    Best answer

    Would that be x = 20t*e^(-1.4t)?

    Set the derivative = 0:

    dx/dt = 20*e^(-1.4t) + 20t*-1.4*e^(-1.4t) = 0

    Then e^(-1.4t) - 1.4te^(-1.4t) = 0

    thus 1.4t = 1 ==> t = 5/7

    I get 5.25542067766274 m at t = 5/7 s.

  • cepero
    Lv 4
    3 years ago

    a) the rate is dx(t)/dt = x'(t), so which you would be able to differentiate x(t). b) resolve the equation x'(t) < 0. this provide you with a situation on x. c) Differentiate x'(t) to get x''(t) and set x''(t) = 0. resolve and take any answer the place x'(t) <> 0. Notes: a million) the region is a 4th-order function with 4 roots a million, a million, a million,a million.5, 2) the rate is a third-order function with 3 roots a million, a million and x 3) The acceleration is a 2d-order function with 2 roots a million, y

  • 30e-31m

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