# Calculus

I really need help with this problem

an automobile accelerates from rest at 1 + 3(square root)t mph/sec for 9 seconds.

How far does it travel in those nine seconds?
I have found that the velocity, after nine seconds, is 63

I have figuredout the answer but i don't know how to do it the answer is 344.52 feet

the derivative of distance gives us velocity, and the derivative of velocity gives us acceleration.

So lets works backwards
if a(t)=1 + 3√t then
v(t) = t + 2t^(3/2) + c
but we were told that when t=0, v=0 (from rest), so c=0

now let d be the distance
then d(t) = .5t^2 + (4/5)t^(5/2) + k

but when t=0, d=0 so k=0

I think the problem is the units.
The original "a" was in mph/sec

If I recall my archaic measurement units, 1 mile was 5280 feet, and of course 1 hour = 3600 seconds
so 1mph = 5280/3600 ft/sec

so if we change our distance formula to read

d(t) = (528/360)(.5t^2 + (4/5)t^(5/2)

we can input seconds and get feet as output
when t=9
d(9) = 528/360*(.5(81) + .8(9)^2.5_
= 344.52 feet

thanks so much I was lost on that the measurement was what threw me off.

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