Question

an automobile accelerates from rest at 1+ 3* sqrt (t) mph/sec for 9 seconds.
how far does it travel in those 9 sec?

Answer 1

Need two integrals. First integral will be from 0 to 9 sec, and will give the speed in MPH as a function of t in seconds.

Answer 2

For the second integral, I think a change of variable is in order to get the units to match.

Answer 3

\[velocity=\int\limits_{}^{}a dt\]\[distance = \int\limits_{}^{} velocity dt\]
so v = t + (3 * t^1.5)/1.5
which is v = t + 2t^1.5

distance = t^2/2 + (2 t^2.5)/2.5
distance = t^2/2 + 4/5t^2.5

make it simpler t^2.5 is same as sqrt(t^5)

distance = t^2/2 + 0.8 * sqrt(t^5)

put in time at end of 9
distance = 40.5 + 0.8 * sqrt(59049)
distance = 40.5 + 0.8 * 243

distance = 234.9
(I think! - I'm a little rusty)

Answer 4

forgot to mention the trick in the first line (sorry)
sqrt(x) is same as x^0.5
to integrate use the "use the standard rule of adding 1 to the current power then dividing by the new power" so it becomes x^(0.5+1)/1.5, so x^1.5/1.5

Answer 5

The calculus seems ok, but you need a fudge factor to correct for the mph/sec in the initial expression for the acceleration. I think the easiest way would be to add a constant factor to change mph to fps. Then when you integrate for the second time, it will give distances in feet.

Answer 6

If I recall correctly, the constant to change mph to fps is 32.2/22 (approx).

Answer 7

you are right - acceleration is quoted in mph. need to convert to m/s = 1609.344m/h = 1609.344/3600 = 0.44704m/s. factor is constant so you can just multiply my answer by the factor - answer is 105m