Velocity function, unit vector notation, acceleration, speed

  • Thread starter ColtonCM
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  • #1
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The problem:

The velocity ~v (vector notation, don't know how to type) is given by:

~v = (6.0t - 4.0t^2)ˆi + 8.0ˆj + (3.0t) with |~v| in meters per second and positive t seconds. ˆi, ˆj, ˆk have their usual meanings (unit vector notation).

(a) What is the acceleration ~a of the particle when t = 3.0 s?
(b) What is the speed of the particle |~v| at t = 3.0 s?
(c) If ever, when would the acceleration of the particle ~a be zero m/s^2?

I know what unit vector notation is and I know that acceleration is the derivative of velocity function and I know how to do the Calculus to obtain those things, but I don't know how what this function is when applied with unit vector notation? i refers to x-axis, j to y-axis, and k to z-axis. Am I allowed to take the derivative of each function to obtain acceleration, do I combine the functions, or in other words how do I handle this question?

Thanks for your time,

Colton Casados-Medve
 

Answers and Replies

  • #2
ehild
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The rules of differentiation hold: the derivative of a sum is the sum of the derivatives, and the derivative of a function multiplied by a constant is the derivative multiplied by the constant. The unit vectors along the axes x, y, z are constants. If ##\vec V = X(t) \hat x + Y(t) \hat y + Z(t) \hat z ##
## \frac{d \vec V}{dt} = \frac{dX(t)}{dt} \hat x+ \frac{dY(t)}{dt} \hat y+ \frac{dZ(t)}{dt} \hat z ##
 
  • #3
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The rules of differentiation hold: the derivative of a sum is the sum of the derivatives, and the derivative of a function multiplied by a constant is the derivative multiplied by the constant. The unit vectors along the axes x, y, z are constants. If ##\vec V = X(t) \hat x + Y(t) \hat y + Z(t) \hat z ##
## \frac{d \vec V}{dt} = \frac{dX(t)}{dt} \hat x+ \frac{dY(t)}{dt} \hat y+ \frac{dZ(t)}{dt} \hat z ##
Got it, thanks a lot!
 

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