What is the average speed of an object that is still at #t=0# and accelerates at a rate of #a(t) =t-3# from #t in [3, 6]#?

2 Answers
Apr 24, 2016

Answer:

#v_a=-3 " unit/s"#

Explanation:

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#v(t)=int(t-3)*d t" "v(t)=t^2/2-3t#

#x(t)=int v(t)*d t" "x(t)=int (t^2/2-3t)*d t#

#x(t)=1/2*t^3/3-3t^2/2" "x(t)=t^3/6-(3t^2)/2#

#"for position t=3 -> x(3)=-9 unit"#

#"for position t=6 ->x(6)=-18 unit"#

# v_a=tan alpha#

#v_a=(-18+9)/(6-3)#

#v_a=-9/3#

#v_a=-3 " unit/s"#

Apr 25, 2016

Answer:

#"Average speed"=-3 " units"#
#-ve# sign indicates that speed is opposite to the defined positive direction of displacement.

Explanation:

Given acceleration #a(t)=t-3#. We know that #a=x''#, therefore, to find distance moved during the period of interest #x=int int a(t).dt.dt# over the period of interest
(First integral will give us the velocity in terms of time.) Inserting given values.

#x=int int(t-3).dt.dt#
#=int_3^6(t^2/2-3t+C).dt# .....(1), where #C# is constant of integration.
We need to evaluate #C# from initial conditions
Velocity #v(t)=t^2/2-3t+C#
Now #v(0)=0^2/2-3xx0+C=0#, initial condition
or #C=0#
Inserting in (1)

#x=int_3^6(t^2/2-3t).dt#
#x=|t^3/6-3 t^2/2|_3^6#, constant of integration ignored due to definite integral

#=(6^3/6-3xx 6^2/2)-(3^3/6-3 xx3^2/2)#
#=36-54-9/2+27/2#
#=-9#,
We know that #"Average speed"=("Total Displacement")/"Time Taken"#
#"Average speed"=-9/(6-3)#
#"Average speed"=-3#