An object's two dimensional velocity is given by #v(t) = ( 5t-t^3 , 4t^2-t)#. What is the object's rate and direction of acceleration at #t=4 #?

Question

1461 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-22T16:34:54

The rate of acceleration is #=53ms^-2# in the direction #=144.2^@# anticlockwise from the x-axis

The acceleration is the derivative of the velocity

#v(t)=(5t-t^3, 4t^2-t)#

#a(t)=v'(t)=(5-3t^2, 8t-1)#

When #t=4#

#a(4)=(-43,31)#

The rate of acceleration is

#||a(4)||=sqrt((-43)^2+(31)^2)=sqrt(2810)=53#

The direction is

#theta=arctan(-31/43)=144.2^@# anticlockwise from the x-axis