The position function of a particle moving along a coordinate line is given, wheresis in feet andtis in seconds. s(t)= t/t^2+4, t>=0

a. Find the velocity and acceleration functions.
b. Find the position, velocity, speed, and acceleration at time t=1.
c. At what times is the particle stopped?
d. When is the particle speeding up? Slowing down?
e. Find the total distance traveled by the particle from time t=0 to time t=5

a) s = (4-t)/(t^2+4)^2,   a(t) = (2t^3-24t)/(t^2+4)^3

b) s = 0.2ft, v = 0.12 ft/s,  a = -0.176 ft/s^2

c) t = 2s

d) slowing down for t < 2, speeding up for t > 2

e) 0.327 ft

Step-by-step explanation:

The position function of a particle is given by:

  (1)

a) The velocity function is the derivative, in time, of the position function:

  (2)

The acceleration is the derivative of the velocity:

(3)

b) For t = 1 you have:

c) The particle stops for v(t)=0. Then you equal equation (2) to zero ans solve the equation for t:

For t = 2s the particle stops.

d) The second derivative evaluated in t=2 give us the concavity of the position function.

Then, the concavity of the position function is negative. For t=2 there is a maximum. Before t=2 the particle is slowing down and after t=2 the particle is speeding up.

e) Due to particle goes and come back. You first calculate s for t=2, then calculate for t=5.

The particle travels 0.25 in the first 2 seconds. In the following three second the particle comes back to the 0.172\ ft. Then, in the second trajectory the particle travels:

0.25 - 0.127 = 0.077 ft

The total distance is the sum of the distance of the two trajectories:

s_total = 0.25 ft + 0.077 ft = 0.327 ft

if x = 1, then the answer is 3.5