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 ?

1 Answer
Sep 17, 2017

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