What is the average speed of an object that is not moving at #t=0# and accelerates at a rate of #a(t) =10-t# on #t in [6, 10]#?

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Answer 1 · 2016-09-20T14:29:34

Given
#"Acceleration"=a(t)=10-t#

#(d(v(t)))/(dt)=10-t#

Integrating we get

#v(t)=int(10-t)dt=10t-1/2t^2+c#

It is also given that the object is not moving at #t=0#,

#"so "v(0)=0=>c=0#

#:.v(t)=10t-1/2t^2#

#=>(d(s(t)))/(dt)=10t-1/2t^2#

Integrating for # t in [6,10]# we get the distance traversed

#s=int_6^10(10t-1/2t^2)dt#

#=[10xxt^2/2-1/6t^3]_6^10#

#=5xx10^2-1/6xx10^3-5xx6^2+1/6xx6^3#

#=536-180-500/3#

#=356-166.67=189.33 #

#"Average speed"=s/(Deltat)=189.33/4~~47.33"unit"/s#