An object's two dimensional velocity is given by #v(t) = ( e^t-2t^2 , t^3 - 4t )#. What is the object's rate and direction of acceleration at #t=7 #?

Question

1312 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-16T05:22:02

The rate of acceleration is #=1078.1ms^-2# in the direction of #7.62^@# anticlockwise from the x-axis

Acceleration is the derivative of the velocity

The velocity is

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

The acceleration is

#a(t)=v'(t)=(e^t-4t,3t^2-4)#

Therefore, when #t=7#

#a(7)=(e^7-28,3*49-4)=(1068.6,143)#

The rate of acceleration is

#||a(7)||=sqrt(1068.6^2+143^2)=1078.1ms^-2#

The direction is

#theta=arctan(143/1068.6)=7.62^@# anticlockwise from the x-axis