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

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) = te^?t/4 (a) Find the velocity at time t (in ft/s). v(t) = (b) What is the velocity after 7 s? (Round your answer to two decimal places.) v(7) = (c) When is the particle at rest? t = (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 10 s. (Round your answer to two decimal places.) (f) Find the acceleration at time t (in ft/s2). a(t) = Find the acceleration after 7 s. (Round your answer to three decimal places.) a(7) = (g) Graph the position, velocity, and acceleration functions for 0 ? t ? 10. (h) When is the particle speeding up? (Enter your answer using interval notation.) (i) When is it slowing down? (Enter your answer using interval notation.)

Explanation / Answer

(a.) f '(t) = v(t) = -t/4e^(-t/4) + e^(-t/4)

(b.) v(7) = -7/4e^(-7/4) + e^(-7/4) = -0.13 ft/s

(d.) Find the times where v(t) > 0.

(e.) %u222B|e^(-t/4)(1 - t/4)| dt from 0 to 10

(f.) a(t) = v '(t)

Plug in 7 into a(t) fo find a(7).

(h) Wherever a(t) > 0, the particle is speeding up.

Wherever a(t) < 0, the particle is slowing down.

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differentiate position with respect to time and you get:
v=ds/dt (velocity)
a=dv/dt (acceleration)
j=da/dt (jerk)

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