speed of bullet

[UrbanXrisis]

The speed of a bullet as it travles down the barrel of a rifle towards the opening is given be the expression v=(-5.0*10^7)t^2 + (3.0*10^5)t, where v is in meters per second and t is in seconds. The acceleration of the bullet just as it leaves the barrel is zero.

(a) determine the acceleration and position of the bullet as a function of time when the bullet is in the barrel.

I multiplied everything by t to get the position equation: x=(-5.0*10^7)t^3 + (3.0*10^5)t^2

I divided everything by t to get the acceleration equation: a=(-5.0*10^7)t + (3.0*10^5)

is this thougth process correct?

(b) determine the length of time the bullet is accelerated.
you dont know the length of the barrel so is this possible?

(c) Find the speed at which the bullet leaves the barrel
based on question b

(d) what is the length of the barrel
based on question b as well

in need of need hits

[Tide]

Differentiate to find the acceleration and integrate to find the position.

[UrbanXrisis]

what does differentiate mean?

[Tide]

Differentiate means finding the derivative. I assumed from the stated problem that you probably have some calculus experience. If not then you may have to resort to graphing and finding the slope of the curve at several points to make a graph of acceleration.

[UrbanXrisis]

for question B, I need to find the time, how would I do that?

[Tom McCurdy]

a= 300000-100000000t
0=300000-100000000t
t=3/1000

x=150000x^2-\frac{50000000x^3}{{3}}
x=.9 meters

[Tide]

For that one you have to integrate!

[Tom McCurdy]

for the velocity question just plug in when you solved for time

[Tom McCurdy]

I just found the derivative of the V for a and integrated V for x

[UrbanXrisis]

how do you know acceleration is zero?

[Tom McCurdy]

Easy way to find derivitave
take each chuck and do dervitiave of cx^n = ncx^{n-1}
to integrate
take
bx^n = (n+1)x= c/(n+1)x^{n+1}

where n is power
c is orignial coeffiecnt
x is variable

[Tom McCurdy]

The speed of a bullet as it travles down the barrel of a rifle towards the opening is given be the expression v=(-5.0*10^7)t^2 + (3.0*10^5)t, where v is in meters per second and t is in seconds. The acceleration of the bullet just as it leaves the barrel is zero.


You gave that to me in the problem

[UrbanXrisis]

oh, that's right!! So the speed of the bullet would just be m/s.....9m/.003s?

[Tom McCurdy]

I just found velocity for it usuing the original equation
you gave me
which gave a velocity of 450 m/s

[UrbanXrisis]

wait...all I have to do is sub .003 into the original velocity equation to get 450m.s right?

[UrbanXrisis]

okay! I get it! Thanks a lot! Could you take a quick look at this other problem which I already figured out but wanted to make sure it was accurate?

http://www.physicsforums.com/showthread.php?p=326372#post326372

[Tom McCurdy]

btw the x distance i got was 0.9 meters not 9 meters