The functions(t)describes the motion of a particle along a line.s(t)=t2−12t+22

Question

(a) Find the velocity functionv(t)of the particle at any time t≥0

v(t)=

(b) Identify the time interval(s) in which the particle is moving in a positive direction. (Enter your answer using interval notation.)

(c) Identify the time interval(s) on which the particle is moving in a negative direction. (Enter your answer using interval notation.)

(d) Identify the time(s) at which the particle changes direction. (Enter your answers as a comma-separated list.)

Answer 1

The approach is to differentiate the given function which is a parabola or second order polynomial. The resulting differentiated function a linea,r y=m*x +b, equation and represents the speed of the object in the x direction.

Set the linear function to less than zero and calculat the correcponding x region which is the region of negative speed. Similarly set the linear function to greater than zero to calculat the correcponding x region which is the region of postive speed.

Note: the speed are in the x direction.

Be well!

Answer 2

s(t) =-16t^2 -12t + 22

velocity is the derivative of distance

v(t) = -32t -12

it moves in a positive direction when v(t)>0

t= -12/32 = -3/8 = -0.375 seconds, when the particle is not moving, for an instant in time

when t>-3/8 it's moving backward

(-3/8, infinity) backward moving

(-infinity, -3/8) forward moving

it changes direction at time t= -3/8 = -.375 second

but you seem to want to restrict t to positive numbers

so that means the interval would be (0, infinity) backward

it changes direction at time t= -3/8 part of a second when it reaches maximum distance 29 7/8 feet

but if you restrict the domain to t>0, the particle never changes direction, as it's always moving backward with positive time

it's distance is graphically a downward opening parabola with vertex (-3/8, 29 7/8)

s(t) =-16(t +3/8)^2 + 29 7/8 is the quadratic in vertex form