An object's two dimensional velocity is given by #v(t) = ( t^2, t-t^2sin(pi/3)t)#. What is the object's rate and direction of acceleration at #t=4 #?

Question

An object's two dimensional velocity is given by #v(t) = ( t^2,  t-t^2sin(pi/3)t)#.  What is the object's rate and direction of acceleration at #t=4  #?

Amelia Adams · Accepted Answer

Please see the explanation.

#v_x(t) = t^2# #(dv_x(t))/dt = 2t# #v_y(t) = t - t^2sin((pit)/3)# #(dv_y(t))/dt = 1 - 2tsin((pit)/3) - (pit^2)/3cos((pit)/3)# The magnitude of the acceleration is #a = sqrt((2t)^2 + (1 - 2tsin((pit)/3) - (pit^2)/3cos((pit)/3))^2) # I used WolframAlpha to evaluate the above at t = 4: #a ~~ 9.43# The direction is along the unit vector: #(1/9.43)(64hati - 4.99hatj)#

Parker Abbott · Answer

undefined To find the acceleration vector, differentiate the velocity vector with respect to time. Then evaluate the resulting expression at t = 4.

v(t) = (t^2, t - t^2sin(π/3)t)

Differentiate each component with respect to time:

a(t) = (2t, 1 - 2tsin(π/3) - t^2cos(π/3))

Evaluate at t = 4:

a(4) = (8, 1 - 8sin(π/3) - 16cos(π/3))

Now, compute the magnitude of the acceleration vector:

|a(4)| = sqrt(8^2 + (1 - 8sin(π/3) - 16cos(π/3))^2)

Finally, find the direction of acceleration by calculating the angle between the acceleration vector and the positive x-axis using:

θ = arctan((1 - 8sin(π/3) - 16cos(π/3)) / 8)

Substitute the values for sin(π/3) and cos(π/3) and compute the angle.