An object's two dimensional velocity is given by #v(t) = ( t^3, t-t^2sin(pi/8)t)#. What is the object's rate and direction of acceleration at #t=4 #?

1 Answer
Jul 10, 2017

Answer:

#a = 48.0# #"m/s"^2#

#theta = 1.19^"o"#

Explanation:

We're asked to find the object's acceleration (magnitude and direction) given its velocity equation.

We need to find the object's acceleration as a function of time, by differentiating the velocity component equations:

#a_x = d/(dt) [t^3] = 3t^2#

#a_y = d/(dt) [t - t^2 sin((pi/8)t)] = 1 - 2t^2cos((pi/8)t)#

Note: I changed the #y#-velocity equation SLIGHLTY by moving the final #t# inside the sine function, because (1) otherwise it could have said #t - t^3sin(pi/8)# and (2) this is how I've always seen these equations.

How that we know the acceleration components as functions of time, let's plug in #t = 4# #"s"# to find the components at that time:

#a_x (4) = 3(4)^2 = 48# #"m/s"^2#

#a_y = 1 - 2(4)^2cos((pi/8)(4)) = 1# #"m/s"^2# (cosine of #pi/2# is #0#)

The magnitude of the acceleration is thus

#a = sqrt((48color(white)(l)"m/s"^2)^2 + (1color(white)(l)"m/s"^2)^2) = color(red)(48.0# #color(red)("m/s"^2#

And the direction is

#theta = arctan((1cancel("m/s"^2))/(48cancel("m/s"^2))) = color(blue)(1.19^"o"#