Monday, October 12, 2015

Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 61

The motion's equation of a particle is s=t33t, where s is in meters and t is in seconds.
a.) Find the velocity and aceleration as function of t
Given: s=t33t
Take the 1st derivative of the given equation to get the velocity and 2nd derivative to get the acceleration.


V(t)=t33tV(t)=ddt(t3)3ddt(t)(Derive each term)V(t)=3t23(1)(Simplify the equation)


The velocity of a particle as function of t is V(t)=3t23


a(t)=3t23a(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)=3t23xa(t)=6t



V(t)=3t23Substitute the given Velocity0=3t23Add 3 to each sides3t2=3Divide both sides by 33t23=33Take the square root of both linest2=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

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