A particle moves according to a law of motion s = f(t), t 0, where t is measured

Question

A particle moves according to a law of motion s = f(t), t 0, where t is measured in seconds and s in feet. f(t) = tet/2 (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 s. (Round your answer to two decimal places.) (f) Find the acceleration at time t (in ft/s2). Find the acceleration after 2 s. (Round your answer to three decimal places.) (h) When, for 0 t < , is the particle speeding up? (Enter your answer using interval notation.) When, for 0 t < , is it slowing down? (Enter your answer using interval notation.)

Explanation / Answer

f(t) = te-t/2

(d) Positive direction = when v(t) > 0
= f'(t) > 0

-1/2 * (e-t/2)*(t-2) >0

(2-t) * (e-t/2) > 0 =>> t<2

Therefore, the particle will move in the positive direction after 2 seconds.

e) f(t) = te-t/2

f(6) = 6 * e-6/2

f(6) = 6e-3 = 0.298722 feet ~ 0.29 feet

f) Acceleration = f''(t) = 1/4 * (e-t/2) * (t-4)

Therefore,

acceleration = ((t-4) * (e-t/2)) / 4

After 2 seconds,

accceleration = ((2-4) * (e-2/2)) / 4 = -e-1 / 2 = 0.183 m/s2

Answers may be wrong because there is no clarity in the expression whether the expression is tet/2 or te-t/2 however, the logic is correct. So you may follow the same just in case the expression used is wrong.

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