A particle moves so that its position vector with respect to the origin of a reference frame Oxyz is r(t)=bcos w t+bsin w t+vt k, where i, j and k are unit vectors parallel to the co-ordinates axes Ox, Oy and b&V are positive constants. (i)Find the velocity and speed of the particle
(ii)Describe the path moved by the particle
(iii)Find the acceleration of the particle

(i) Velocity is the rate of change of position, so if

r(t) = b cos(ω t ) i + b sin(ω t ) j + v t k

then

v(t) = dr/dt

v(t) = -b ω sin(ω t ) i + b ω cos(ω t ) j + v k

The speed of the particle is the magnitude of the velocity, given by

|| v(t) || = √[(-b ω sin(ω t ))² + (b ω cos(ω t ))² + v ²]

… = √[b ²ω ² + v ²]

(ii) The path is a helix. Suppose you zero out the k component. Then the path is a circle of radius b, and the value of ω determines how quickly a particle on the path traverses the circle. Now if you reintroduce the k component, the value of v will determine how far from the plane z = 0 the particle moves in a helical path as t varies.

(iii) Acceleration is the rate of change of velocity, so

a(t) = dv/dt

a(t) = -b ω ² cos(ω t ) i - b ω ² sin(ω t ) j

f(x) has value 150 at x= 0

and 0 at x= 3

so d(f(x))/dt = slope = -150/3 = -50

-7/8

step-by-step explanation:

instead of subtracting you add the -6/8 and -1/8