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Part $1$

A hot$-$air balloon $is$

$140ft140ft$

above the ground when a motorcycle $(travelingin$ a straight line $on$ a horizontal road$)$ passes directly beneath $it$ going

$30mi$ divided $byhr30m\frac{i}{h}r$

$(44ft$ divided $bys44f\frac{t}{s}).$

$If$ the balloon rises vertically $at$ a rate $of$

$6ft$ divided $bys6f\frac{t}{s},$

what $is$ the rate $of$ change $of$ the distance between the motorcycle and the balloon

$8$ seconds$8$ seconds

later?

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Part $1$

The rate $of$ change $of$ the distance between the motorcycle and the balloon after

$8$ seconds$8$ seconds

$is$ about

enter your response here

$ft$ divided $bys.f\frac{t}{s}.$

$(Roundto$ two decimal places $as$ needed.$)$