Question

A particle has a position function →r(t)=cos(1.0t)ˆi+sin(1.0t)ˆj+tˆk, where the arguments of the cosine and sine functions are in radians. (a) What is the velocity vector? (b) What is the acceleration vector?

a) (i) v(t)=-sin(t)ˆi+cos(t)ˆj+ˆk m/s
(ii) v(t)=-sin(t)ˆi+cos(t)ˆj+1ˆk m/s
(iii) v(t)=-cos(t)ˆi-sin(t)ˆj+1ˆk m/s
(iv) v(t)=-cos(t)ˆi-sin(t)ˆj+ˆk m/s

b) (i) a(t)=-cos(t)ˆi-sin(t)ˆj m/s²
(ii) a(t)=-cos(t)ˆi-sin(t)ˆj+ˆk m/s²
(iii) a(t)=-sin(t)ˆi+cos(t)ˆj m/s²
(iv) a(t)=-sin(t)ˆi+cos(t)ˆj+ˆk m/s²

Answer 1

The velocity vector is -sin(t)i + cos(t)j + k m/s and the acceleration vector is -cos(t)i - sin(t)j m/s².

Step-by-step explanation:

The velocity vector can be found by taking the derivative of the position function. In this case, the velocity vector is given by:

v(t) = -sin(t)i + cos(t)j + k m/s

The acceleration vector can be found by taking the derivative of the velocity function. In this case, the acceleration vector is given by:

a(t) = -cos(t)i - sin(t)j m/s²