Question

The position function of a particle moving along a coordinate line is given, wheresis in feet andtis in seconds.

Answer 1

Answer:

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:

$s(t)=\frac{t}{t^2+4},\ \ \ t\geq 0$   (1)

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

$v(t)=\frac{ds}{dt}=\frac{(1)(t^2+4)-t(2t)}{(t^2+4)^2}=\frac{4-t^2}{(t^2+4)^2}$   (2)

The acceleration is the derivative of the velocity:

$a(t)=\frac{dv}{dt}=\frac{(-2t)(t^2+4t)^2-(4-t^2)2(t^2+4)(2t)}{(t^2+4)^4}\\\\a(t)=\frac{(-2t)(t^2+4)-4t(4-t^2)}{(t^2+4)^3}=\frac{2t^3-24t}{(t^2+4)^3}$ (3)

b) For t = 1 you have:

$s(1)=\frac{1}{1+4}=0.2\ ft\\\\v(1)=\frac{4-1}{(1+4)^2}=0.12\frac{ft}{s}\\\\a(1)=\frac{2-24}{(1+4)^3}=-0.176\frac{ft}{s^2}$

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

$v(t)=\frac{4-t^2}{(t^2+4)^2}=0\\\\4-t^2=0\\\\t=2$

For t = 2s the particle stops.

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

$\frac{d^2s}{dt^2}=a(2)=\frac{2(2)^3-24(2)}{(2^2+4)^3}=-0.062$

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.

$s(2)=\frac{2}{2^2+4}=0.25\ ft$

$s(5)=\frac{5}{5^2+4}=0.172\ ft$

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