An object's velocity is given by v(t) = (2t^2 +t +1 , sin2t ). What is the object's rate and direction of acceleration at t=6 ?

1 Answer
Nathan L.
Jul 16, 2017

a = 27.1 "LT"^-2

theta = 3.58^"o"

Explanation:

We're asked to find the the magnitude (rate) and direction of an object's acceleration at a given time, given its velocity equation.

To do this, we need too differentiate the velocity component equations, which are given as

v_x(t) = 2t^2 + t + 1

v_y(t) = sin(2t)

Finding the derivatives:

a_x(t) = d/(dx) [2t^2 + t + 1] = 4t + 1

a_y(t) = d/(dx)[sin(2t)] = 2cos(2t) (I'll assume this is in radians**)

Plugging in t = 6 (no units), we have

a_x = 4(6) + 1 = 27 "LT"^-2

a_y = 2cos(2(6)) = 1.69 "LT"^-2

(The "LT"^-2 term is the dimensional form of the units for acceleration ("distance"xx"time"^-2). I used this term here since no units were given.)

The magnitude of the acceleration is thus

a = sqrt((a_x)^2 + (a_y)^2) = sqrt(27^2 + 1.69^2) = color(red)(27.1 color(red)("LT"^-2

And the direction is

theta = arctan((a_y)/(a_x)) = arctan(1.69/27) = color(blue)(3.58^"o"

Always make sure your arctangent calculation is in the right direction; it could be 180^"o" off!

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