A particle moves according to a law of motion s = f(t), t2 0, where t is measured in seconds and s in feet.f(t) = 0.01t4 - 0.03t3(a) Find the velocity at time t (in ft/s).v(t) =%3D(b) What is the velocity after 1 second(s)?v(1) =ft/s%3D(c) When is the particle at rest?smaller valuet =S.larger value(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 9 seconds. (Round your answer to two decimal places.)(f) Find the acceleration at time t (in ft/s2).a(t) =Find the acceleration after 1 second(s).a(1) =ft/s2ae Type here to search

Given particle moves in s = 0.01t4-0.03t3 a) Hence the velocity of the particle is v = dsdt = 0.04…

Answered: A particle moves according to a law of…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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A golf ball is fired straight up into the air from ground level. It reaches a maximum height at ℎ = 504 meters. The speed of the golf ball in meters per second is 99.41. Calculate the flight time in ttotal in seconds until the golf ball reaches the ground.
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A person stands at the edge of a cliff and throws a rock horizontally over the edge with a speed of v = 23.0 m/s. The rock leaves his hand at a height of h = 43.0 m above level ground at the bottom of the cliff, as shown in the figure. Note the coordinate system in the figure, where the origin is at the bottom of the cliff, directly below where the rock leaves the hand. 4 7 i (a) What are the coordinates of the initial position of the rock? (Enter your answers in m.) Xo = Yo = m (b) What are the components of the initial velocity? (Enter your answers in m/s.) Vox = m/s m/s Voy
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The acceleration of a particle is a constant. At t = 0 the velocity of the particle is (15.8î + 18.4ĵ) m/s.At = 3.6 s the velocity is 10.5ĵ m/s. (Use the following as necessary: t. Do not include units in your answers.) (b) How do the position (in m) and velocity (in m/s) vary with time? Assume the particle is initially at the origin.
At time t = 0 a particle starts moving along a line with equation of motion s(t) = t^3−4t^2+ 6, whereS is meters and t is in seconds.a) Give the velocity and acceleration functions.b) What is the velocity at time t = 2?c) When is the acceleration 2 m/s^2?
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