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 − 8t2 + 25t

Question

Answered:

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(a) Find the velocity at time t

A) 3t^2 - 16t + 25

(b) What is the velocity after 1 second?

A) 12

(c) When is the particle at rest?

A) DNE

(g) Find the acceleration at time t and after 1 second

A) a(t) = 6t - 16, a(1) = -10

Needed:

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

(e) Find the total distance traveled during the first 6 seconds.

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

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

Answer 1

d) Since your velocity equation 3t^2 - 16t +25 is always positive, the particle is ALWAYS moving in the positive direction! Interval notation would be (0, + infinity)

e) Since the particle is always moving in the positive direction, the total distance traveled is the same as the displacement. So s(6) - s(0) = 78 - 0 = 78 feet. Alternatively, we could use calculus to solve this problem by taking the definite integral of v(t) from 0 to 6. The result is the same.

i) This is asking for when the acceleration is positive. Using your equation a(t) = 6t - 16, we can see that a(t) goes from negative to positive at t = 8/3. So the interval would be (8/3, +infinity)

ii) Using the same reasoning as part (i), the acceleration is negative on the interval (0, 8/3)

Let me know if any part of this explanation needs clarification!

Answer 2

(a) v(t) = ds/dt = 3t² - 16t + 25

(b) v(1) = 3 - 16 + 25 = 12 ft/s

(c) v(t) = 3t² - 16t + 25 = 0, real so.lution DNE, so, particle never stop

(d) a(t) = dv/dt = 6t - 16; a(1) = 6 - 16 = - 10 ft/s²