I don't understand the reasoning behind a few steps in this problem:
What constant acceleration is required to increase the speed of a car from 30 mi to 50 mi in 5 seconds?
a(t) = C = v'(t)
v'(t) = Ct + K
v(0)=30= C(0) + k
k = 44 ?
v(5) = (5/6)88 ft/s ?
v(5) = 5c + 44
5c = 88/3 = 88/15 ft/s^2
I think I know where k = 44 comes from.
50 mi - 30mi = 20mi
20mi/5s = 4mi/s(30mi) = 120mi/s
5280/120 = 44
But I don't understand where the other number comes from.
Lets assume the change in speed is from 30 mph to 50 mph, that is a change
of 20 mph. Convert this to fps:
dv=[20*5280]/[60*60] = 29.333.. fps
So the speed changes by 29.333.. fps in 5 seconds so the avaerage
acceleration is 29.333../5 ~= 5.866.. f/s^2
Which is the constant acceleration that will change the speed by the
required amount in 5 seconds.
RonL
You seem to have v(0) = 30 = C.0 + k = k
so in these units k=30.
But watch your units, in you next section you are using fps, while here you
have mi (whatever they are).
Now if we convert 30 mph to fps we have 44 fps, and if we are working in fps
we have:
v(0) = 44 = C.0 + k = k
so in fps k does equal 44 fps.
RonL
I made a mistake it my first post, it should have been mi/h (mph). Anyway, I'm sure you know what it should be.But watch your units, in you next section you are using fps, while here you
have mi (whatever they are).
Captin, I know that it should equal 44 but I don't know why. Also, you're not answering my other question as I still don't know why v(5) = 440/6 and it doesn't look like I'm going to get any help with that.Now if we convert 30 mph to fps we have 44 fps, and if we are working in fps
we have:
v(0) = 44 = C.0 + k = k
so in fps k does equal 44 fps.
Anyway, thanks . . .
Can someone else explain how to convert from mph to fps, completely, and by using fractions only?
Ok, i hope you get this.
first let's do the conversions:
1 mile = 5280 ft
1 hour = 3600 sec
...30 miles...........5280 ft...........1 hour..........30(5280) ft
.-----------..x..-----------..x..----------..=..-------------..= 44 ft/s
.....hour..............1 mile...........3600 sec..........3600 sec
did you get that method? i used conversion factors that you know of, and i placed them in such a way, that the units i don't want cancel out, and the units i do want stay. (i hate this method, but people seem to understand it easier). similarly,
...50 miles...........5280 ft...........1 hour..........50(5280) ft
.-----------..x..-----------..x..----------..=..-------------..= 220/3 ft/s
.....hour..............1 mile...........3600 sec.........3600 sec
Now let's get to the meat of the matter. we now know 30 mph = 44 ft/s, and 50 mph = 220/3 ft/s
a(t) = C = v'(t)
=> v(t) = int{a(t)}dt = int{C}dt = Ct + D
so v(t) = Ct + D
at t = 0, v(t) = 44 ft/s
=> 44 = C(0) + D
=> 44 = D
so v(t) = Ct + 44
Now, when t = 5, we attain 220/3 ft/s
=> 220/3 = C(5) + 44
=> 5C = 220/3 - 44 = 88/5
=> C = (88/5)/5 = 88/15
so v(t) = (88/15)t + 44
thus the constant of acceleration is 88/15
if you have problems with my conversion procedure i can show you another way, the one i usually use when i'm writing stuff out
it won't hurt to show you the other method anyway. this method i think is a bit simpler. we simply replace what we don't want with what we do want.
1 mile = 5280 ft
1 hour = 3600 sec
this means wherever i see mile i can replace it with 5280 ft, becuase they are equal, they're the same thing. and wherever i see hour i can replace that with 3600 sec since they're the same thing. so,
30 miles/hour = 30 (5280 ft)/(3600 sec) = 30(5280)/3600 ft/sec = 44 ft/sec