An object's two dimensional velocity is given by v(t) = ( sin(pi/3t) , 2cos(pi/2t )- 3t )v(t)=(sin(π3t),2cos(π2t)3t). What is the object's rate and direction of acceleration at t=1 t=1?

1 Answer
Sep 17, 2017

a=sqrt(pi^2/36+16)a=π236+16

Explanation:

v_x(t)=sin(pi/3t)vx(t)=sin(π3t) and v_y(t)=2cos(pi/2t)-3tvy(t)=2cos(π2t)3t
acceleration in X-direction
a_x(t)=d/dt(v_x(t))=d/dt(sin(pi/3t))=pi/3cos(pi/3t)ax(t)=ddt(vx(t))=ddt(sin(π3t))=π3cos(π3t)

acceleration in Y-direction
a_y(t)=d/dt(v_y(t))=d/dt(2cos(pi/2t)-3t)ay(t)=ddt(vy(t))=ddt(2cos(π2t)3t)
a_y(t)=-2(pi/2)sin(pi/2t)-3=-pisin(pi/2 t)-3ay(t)=2(π2)sin(π2t)3=πsin(π2t)3

at t=1 sec
a_x(1)=pi/3cos(pi/3(1))=pi/3 cos(pi/3)=pi/3(1/2)=pi/6ax(1)=π3cos(π3(1))=π3cos(π3)=π3(12)=π6
a_y(1)=-pisin(pi/2 (1))-3=-pisin(pi/2)-3=-pi-3ay(1)=πsin(π2(1))3=πsin(π2)3=π3
a_y(1)=-pi-3ay(1)=π3

Hence Acceleration a=sqrt((a_x(t))^2+(a_y(t))^2a=(ax(t))2+(ay(t))2
a=sqrt((pi/6)^2+(-pi-3)^2)=sqrt(pi^2/36+pi^2+6pi+9)approx6.1638 a=(π6)2+(π3)2=π236+π2+6π+96.1638

direction is tan^-1(a_y/a_x)=tan^-1((-pi-3)/(pi/6))approx-85.13^o tan1(ayax)=tan1(π3π6)85.13o from Positive X- axis