#color(blue)(s=t^3+3/2t^2-6t+12)#
Part a - To find average velocity: Find average slope of position function:
#frac{s(5)-s(0)}{5-0}#
#=frac{(125+75/2-30+12)-12}{5}#
#=26.5# m/s
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Part b - To find velocity at #t=3#: Find velocity function by differentiating the position function, and plug in #t=3#
#v(t)=3t^2+3t-6#
#color(blue)(v(t)=3(t^2+t-2))#
#v(3)=3(3^2+3-2)=30# m/s
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Part c - Avg acceleration is the average slope on the interval #t in [5,6]#. ("during the 5th second" refers to the time interval from 5 to 6)
#frac{v(6)-v(5)}{6-5}#
#=3(6^2+6-2)-3(5^2+5-2)#
#=36# m/s/s
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Part d - We have to find where the velocity is negative, because that means the particle is moving in a negative direction. Set #v(t)# equal to zero, and determine intervals between those values that are negative.
#0=v(t)=3(t^2+t-2)#
#0=3(t-1)(t+2)#
#t ne -2,t=1#
After testing values on the interval #t in (0,1)# and #t in (1,oo)#, I found that #v(t)# is negative only when #t in (0,1)#, meaning #0 < t < 1#
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Part e - Find #a(3)# by first differentiating #v(t)# to get #a(t)#:
#color(blue)(a(t)=3(2t+1))#
#a(3)=3(7)=21#
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Part f - Find where particle is at rest by setting velocity equal to zero, which we already did in part d. We found that #v(1)=0#, so now we need to find where the particle is at #t=1# by plugging it in to the position function:
#s(1)=(1)^3+3/2(1)^2-6(1)+12=1+3/2-6+12=17/2#
#=8.5# meters
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Part g - Find when #a(t)# is equal to zero, then plug that time into #v(t)#
#0=3(2t+1)#
#2t+1=0#
#t=-1/2#
There is no solution because #t<0#, so the acceleration function is never #0# because of the specified interval for #t# in the problem.
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Part h - Set #12# equal to #v(t)#
#12=3t^2+3t-6#
#0=3t^2+3t-18#
#0=t^2+t-6#
#0=(t-2)(t+3)#
#t ne -3, t=2#
Velocity is #12# m/s when #t=2# seconds
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Part i - Graph of particle:
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Part j - To find distance traveled in first 4 seconds, look at graph above and add:
#"(distance from t=0 to t=2) "+" (distance from t=2 to t=4)"#
#=|12-2|+|2-28|#
#=10+26#
#=36# meters
