The motion's equation of a particle is s=t3−3t, where s is in meters and t is in seconds.
a.) Find the velocity and aceleration as function of t
Given: s=t3−3t
Take the 1st derivative of the given equation to get the velocity and 2nd derivative to get the acceleration.
V(t)=t3−3tV′(t)=ddt(t3)−3ddt(t)(Derive each term)V′(t)=3t2−3(1)(Simplify the equation)
The velocity of a particle as function of t is V′(t)=3t2−3
a(t)=3t2−3a′(t)=3ddt(t2)−ddt(3)a′(t)=(3)(2t)−0
The acceleration of a particle as function of t is a′(t)=6t
b.) Find the acceleration after 2s
Given: a(t)=6tt=2 sec.
a(t)=6tUse the formula of acceleration in part(a)a(2)=6(2)Substitute the given time
The acceleration after 2s is a=12ms2
c.) Find the acceleration when the velocity is 0.
Given: xV(t)=0Equation in part(a):xV(t)=3t2−3xa(t)=6t
V(t)=3t2−3Substitute the given Velocity0=3t2−3Add 3 to each sides3t2=3Divide both sides by 33t23=33Take the square root of both lines√t2=√1Simplify the equationt=1Time when Velocity is 0a(t)=6tSubstitute the computed time when Velocity is 0a(1)=6(1)Simplify the equation
When the velocity is 0, the value of the acceleration is a=6ms2
Monday, October 12, 2015
Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 61
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