A particle moves according to a law of motion s = f(t), t ≥ 0, where t is measured in seconds and s in feet. (If an answer does not exist, enter DNE.) f(t) = t^3 − 8t^2 + 28t

a. Find the velocity at time t.
b. What is the velocity after 1 second?
c. When is the particle at rest?
d. When is the particle moving in the positive direction?
e. Find the total distance traveled during the first 6 seconds.

(a) The velocity at time t, is 3t² - 16t + 28.

(b) The velocity after 1 second is 15 ft/s.

(c) The time when the particle is at rest does not exist.

(d) the particle is always moving in positive direction.

(e) The total distance traveled during first 6 seconds is 96 feet.

The given equation of motion;

f(t) = t³ - 8t² + 28t

(a) The velocity at time t, is calculated as follows;

(b) The velocity after 1 second is calculated as follows;

v = 3(1)² - 16(1) + 28

v = 15 ft/s

(c) When the particles is at rest, the velocity is zero;

3t² -16t + 28 = 0

solve the quadratic equation using formula method;

a = 3, b = -16, c = 28

Thus, the time when the particle is at rest does not exist.

(d) The time of motion of the is greater than or equal to zero.

t ≥ 0

when t = 0

f(0) = (0)³ - 8(0)² + 28(0)

f(0) = 0 - 0 + 0

f(0) = 0

when t = 1

f(1) = (1)³ - 8(1) + 28(1)

f(1) = 1 - 8 + 28

f(1) = 21 feet

Thus, the particle started from zero origin and moves towards positive direction.

(e) The total distance traveled during first 6 seconds is calculated as follows;

f(6) = (6)³ - 8(6)² + 28(6)

f(6) = 96 feet.

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a)v = 3 t² - 16 t + 28 , b)    v = 15 ft / s ,  c) the speed is never zero ,  

d) the particle always moves in the positive direction , e)  s = 96 ft

Explanation:

a) For this exercise how should we use the definition of speed

            v = ds / dt

let's make the derivative

           v = 3 t² - 16 t + 28

b) we calculate with t = 1 s

          v = 3 1² - 16 1 +28

         v = 15 ft / s

c) the particle is at rest when the velocity is zero

         0 = 3 t² - 16 t + 28

         3t² - 16t + 28 = 0

we solve the quadratic equation

       t = [16 ±√ (16² - 4 3 28)] / (2 3)

       t = [5.33 ±√ (256 - 336)] / 6

the square root of a negative number is not defined by which the speed is never zero

d) the particle moves in the positive directional when y> 0, therefore

            d (t)> 0

            t³ - 8 t² + 28 t> 0

            t (t² - 8t + 28)> 0

the expression is true for

          t> 0

          t² - 8t +28> 0

we solve the quadratic equation

         t = [8 ±√ (8² - 4 28)] / 2

         t = [8 ±√ (64 -112)] / 2

the square root of a negative number is not defined, therefore the expression never becomes zero, it is always positive

In short the particle always moves in the positive direction

e) we calculate the distance even t = 6 s

           s = 6³ - 8 6² +28 6

           s = 96 ft