An object's two dimensional velocity is given by v(t) = ( e^t-2t , 2t-4e^2 )v(t)=(et2t,2t4e2). What is the object's rate and direction of acceleration at t=a t=a?

1 Answer
May 24, 2017

The rate of acceleration is =sqrt(e^(2a)-4e^(a)+8)ms^-2=e2a4ea+8ms2 in the direction of arctan(2/(e^(a)-2))arctan(2ea2)

Explanation:

The acceleration is the derivative of the velocity.

v(t)=(e^t-2t, 2t-4e^2)v(t)=(et2t,2t4e2)

A(t)=v'(t)=(e^t-2, 2)

Therefore,

A(a)=(e^(a)-2, 2)

The rate of acceleration is

||a(4)||=sqrt((e^(a)-2)^2+2^2)

=sqrt(e^(2a)-4e^(a)+4+4)

=sqrt(e^(2a)-4e^(a)+8)ms^-2

The direction is

theta=arctan(2/(e^(a)-2))