section 3.7 calculus 1550

Question

A graphing calculator is recommended.

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) = t³ − 9t² + 24t

(a)

Find the velocity (in ft/s) at time t.

v(t) = 3t2−18t+24 ft/s

(b)

What is the velocity (in ft/s) after 1 second?

v(1) = 9ft/s

(c)

When is the particle at rest? (Enter your answers as a comma-separated list.)

t = (4,2)

 

 d)

When is the particle moving in the positive direction? (Enter your answer using interval notation.)

______

 

(e)

Draw a diagram to illustrate the motion of the particle and use it to find the total distance (in ft) traveled during the first 6 seconds.

______ft

(f)

Find the acceleration (in ft/s²) at time t.

a(t) = 6t−18 ft/s²

Find the acceleration (in ft/s²) after 1 second.

a(1) =−12 ft/s²

(h)

When is the particle speeding up? (Enter your answer using interval notation.)

 

When is it slowing down? (Enter your answer using interval notation.)

Answer 1

d) When is the particle moving in the positive direction?

The particle moves in a positive direction when the velocity is positive. So solve:

3t² − 18t + 24 > 0

t² − 6t + 8 > 0

(t - 2)(t - 4) > 0

solution 1: t > 0 but t < 2 or [0, 2)

solution 2: t > 4 or (4, ∞)

e) Draw a diagram to illustrate the motion of the particle and use it to find the total distance (in ft) traveled during the first 6 seconds.

Total distance: 16 + 4 + 2 + 2 + 4 + 16 = 44 feet

h) When is the particle speeding up?

I really don't like this type of question because the wording is a bit tricky. Speed is an absolute value so speeding up can occur when the velocity is positive and the acceleration is positive and it can also occur when the velocity is negative and the acceleration is negative. If we assume that is the meaning then here are the times for positive and negative velocity and acceleration:

t V A

0 - 2 + -

2 - 3 - -

3 -4 - +

4 - ∞ + +

So the particle is slowing down between (0, 2) and (3, 4) and speeding up for (2, 3) and (4, ∞)

Answer 2

d) The particle is moving in the positive direction when its velocity is positive. To find the time interval(s) when that happens, you need to look for the values solving v(t)>0.

Here, v(t) = 3t2−18t+24 >0 ⇒ 0 ≤ t ≤ 2 or t≥4

e) The total distance traveled by the particle can be computed by looking for maxima and minima of s(t) over the time interval t ∈ [0,6]. We do that by computing the derivative (i.e. v(t)) and study how its sign changes over the interval. t = 2 is a maximum and t = 4 is a minimum point. We have 3 regions:

t ∈ [0,2]: s(t) increases from s = 0 to s = 20 (distance traveled = 20)

t ∈ [2,4]: s(t) decreases from s = 20 to s = 16 (distrance traveled = 4)

t ∈ [4,6]: s(t) increases from s = 16 to s = 36 (distance traveles = 20)

Overall, the total distance traveled is the sum = 44 (fts)

f) and (g) The particle speeds up (down) when the acceleration is positive (negative), a(t) > 0 (or a(t) <0).

Solving for t returns, speed up when t > 3 s, slowdown otherwise.