Question 732945
Average speed is ALWAYS ( change in total distance ) / ( change in time )
Let the speed of the truck = {{{ s}}} mi/hr
{{{ s + 10 }}} = the speed of the car
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Note that 20 min = {{{ 20/60 = 1/3 }}} hr
The truck gets a head start of {{{ d[1] = s*(1/3) }}}
Start a stopwatch when the car leaves. Let {{{ t }}} =
the time in hrs on the stopwatch when the car catches the truck.
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Equation for the car:
(1) {{{ 100 = ( s + 10 )*t }}}
Equation for the truck:
(2) {{{ 100 - (1/3)*s = s*t }}}
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(1) {{{ 100 = s*t + 10t }}}
(1) {{{ s*t = 100 - 10t }}}
(1) {{{ s = 100/t - 10 }}}
and
(2) {{{ 300 - s = 3s*t }}}
(2) {{{ 3s*t + s = 300 }}}
(2) {{{ s*( 3t + 1 ) = 300 }}}
Substitute (1) into (2)
(2) {{{ ( 100/t - 10 )*( 3t + 1 ) = 300 }}}
(2) {{{ 10*( 10/t - 1 )*( 3t + 1 ) = 300 }}}
(2) {{{ ( 10/t - 1 )*( 3t + 1 ) = 30 }}}
(2) {{{ 30 - 3t + 10/t - 1 = 30 }}}
(2) {{{ -3t - 1 = -10/t }}}
Multiply both sides by {{{ t }}}
(2) {{{ -3t^2 - t = -10 }}}
(2) {{{ 3t^2 + t - 10 = 0 }}}
Using the quadratic formula:
{{{ t = ( -b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 3 }}}
{{{ b = 1 }}}
{{{ c = -10 }}}
{{{ t = ( -1 +- sqrt( 1^2 - 4*3*(-10) )) / (2*3) }}}
{{{ t = ( -1 +- sqrt( 1 + 120 )) / 6 }}}
{{{ t = ( -1 +- sqrt( 121 )) / 6 }}}
{{{ t = ( -1 + 11 ) / 6 }}}
{{{ t = 5/3 }}} hrs
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The car goes 100 mi in {{{ 5/3 }}} hr
Ave speed = {{{ 100 / (5/3)  = 300/5 }}}
{{{ 300/5 = 60 }}} 
The car averages 60 mi/hr
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The truck goes {{{ 100 }}} mi in {{{ 1/3 + 5/3 = 2 }}} hrs
( Note that the 100 mi has the head start included )
Equation for truck:
{{{ 100 = s*2 }}}
{{{ s = 50 }}}
The truck averages 50 mi/hr
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check:
Equation for the car:
(1) {{{ 100 = ( s + 10 )*t }}}
(1) {{{ 100 = ( 50 + 10)*t }}}
(1) {{{ t = 100/60 }}}
(1) {{{ t = 5/3 }}}
Equation for the truck:
(2) {{{ 100 - (1/3)*s = s*t }}}
(2) {{{ 100 - (1/3)*50 = 50*t }}}
(2) {{{ 300 - 50 = 150t }}}
(2) {{{ 150t = 250 }}}
(2) {{{ t = 250/150 }}}
(2) {{{ t = 5/3 }}}
OK