Question

An object's two dimensional velocity is given by `v(t) = ( sin(pi/3t) , 2cos(pi/2t )- 3t )`. What is the object's rate and direction of acceleration at `t=1`?

Answer 1

`a=sqrt(pi^2/36+16)`

Explanation:

`v_x(t)=sin(pi/3t)` and `v_y(t)=2cos(pi/2t)-3t`
acceleration in X-direction
`a_x(t)=d/dt(v_x(t))=d/dt(sin(pi/3t))=pi/3cos(pi/3t)`

acceleration in Y-direction
`a_y(t)=d/dt(v_y(t))=d/dt(2cos(pi/2t)-3t)`
`a_y(t)=-2(pi/2)sin(pi/2t)-3=-pisin(pi/2 t)-3`

at t=1 sec
`a_x(1)=pi/3cos(pi/3(1))=pi/3 cos(pi/3)=pi/3(1/2)=pi/6`
`a_y(1)=-pisin(pi/2 (1))-3=-pisin(pi/2)-3=-pi-3`
`a_y(1)=-pi-3`

Hence Acceleration `a=sqrt((a_x(t))^2+(a_y(t))^2`
`a=sqrt((pi/6)^2+(-pi-3)^2)=sqrt(pi^2/36+pi^2+6pi+9)approx6.1638`

direction is `tan^-1(a_y/a_x)=tan^-1((-pi-3)/(pi/6))approx-85.13^o` from Positive X- axis