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andrew-mc [135]

3 years ago

13

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.

1 answer:

quester [9]3 years ago

5 0

Answer:

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

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The displacement in the afternoon is $S _{afternoon}= (1300\ m ,\ west) + (2500 \ m ,\ north) - (300\ m ,\ vertical)$

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