# Calculus, antiderivatives

A student accelerates from rest at a rate of 3 miles/min^2. How far will the car have traveled at the moment it reaches a velocity of 65 miles/60 min?

a(t) = 3 miles/min^2
so v(t) = 3t + k miles/min
but when t=0, v=0 so k=0
then v(t)=3t
d(t) = 3/2 t^2 + c
but when t=0 d, the distance, equals zero
so c=0

so you have a(t)=3, v(t)=3t, d(t)=3/2 t^2

when vel=65 miles/60minutes is the same as 13/12 miles/min
then 3t=13/12
t=13/36 minutes

at that t, d = 3/2 (13/12)^2 = 1.76 miles

by the way, since 1 mile/min = 60 mph, after 2 seconds, the velocity would have been 6 miles/min or 180 mph, after 3 seconds 9 miles/min or 540 mph.
Where do I invest in this "car", I want to enter it in a Formula I race.

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