I've got a cute problem:
A car starts from rest and moves along a straight line with an acceleration of a = 3* s^(-1/3) (m/s^2) where s is in meters. Determine the car's velocity and position when t = 6 s.
Ok, so here's what I've done:
v dv = a ds
{ v dv = 3 { s^(-1/3) ds { = integral sign
integrated from 0-v and 0-s respectively
1/2 v^2 = 9/2 * s^(2/3) ---> v^2 = 3 * s^(2/3)
v = 3 * s^(1/3)
Now v = ds/dt therefore ds/dt = 3 * s^(1/3)
ds = 3 * s^(1/3) dt ---> ds / ( s^(1/3) ) = 3 dt
s^(-1/3) ds = 3 dt
{ s^(-1/3) ds = 3 { dt (integrated from 0-s and 0-t respectively)
3/2 s^(2/3) = 3 * t ----> s^(2/3) = 2t
s = (2t)^(3/2)
OK, so plug in t=6 s and find that s = 41.57 meters. This is the correct answer.
Now, I want to find out the velocity at t=6. I would assume I could now take the derivative of s(t) to get v(t):
v = ds/dt = d/dt (2 * t^(3/2) ) ----> 2 * (3/2) * t^(1/2)
v = 3 * t^(1/2)
Plug in t=6: v(6) = 3 * 6^.5 = 7.35m/s This is the wrong answer...
Now if I reuse the original velocity equation (in terms of position s):
v(s) = 3 * s ^(1/3) ---> v(41.57) = 3 * (41.57)^(1/3)
V = 10.39 m/s -- This is the correct answer.
So why doesn't taking the derivative work?
A car starts from rest and moves along a straight line with an acceleration of a = 3* s^(-1/3) (m/s^2) where s is in meters. Determine the car's velocity and position when t = 6 s.
Ok, so here's what I've done:
v dv = a ds
{ v dv = 3 { s^(-1/3) ds { = integral sign
integrated from 0-v and 0-s respectively
1/2 v^2 = 9/2 * s^(2/3) ---> v^2 = 3 * s^(2/3)
v = 3 * s^(1/3)
Now v = ds/dt therefore ds/dt = 3 * s^(1/3)
ds = 3 * s^(1/3) dt ---> ds / ( s^(1/3) ) = 3 dt
s^(-1/3) ds = 3 dt
{ s^(-1/3) ds = 3 { dt (integrated from 0-s and 0-t respectively)
3/2 s^(2/3) = 3 * t ----> s^(2/3) = 2t
s = (2t)^(3/2)
OK, so plug in t=6 s and find that s = 41.57 meters. This is the correct answer.
Now, I want to find out the velocity at t=6. I would assume I could now take the derivative of s(t) to get v(t):
v = ds/dt = d/dt (2 * t^(3/2) ) ----> 2 * (3/2) * t^(1/2)
v = 3 * t^(1/2)
Plug in t=6: v(6) = 3 * 6^.5 = 7.35m/s This is the wrong answer...
Now if I reuse the original velocity equation (in terms of position s):
v(s) = 3 * s ^(1/3) ---> v(41.57) = 3 * (41.57)^(1/3)
V = 10.39 m/s -- This is the correct answer.
So why doesn't taking the derivative work?
