0 </= t </= 9
dx/dt = v = 1 sin (pi t/4)
Now I suspect a typo and you mean
for 0 </= t </= 9
when is v positive ?
from t = 0 to pi t/4 = pi
or t = 0 to t = 4 v is +
from t = 4 to t = 8 v is negative
from t = 8 to t = 9 we are starting the next cycle and t is positive
NOW, the rest of the story (I am sure there is a part B)
If
dx/dt = 1 sin (pi t/4)
then
x = - (4/pi)cos (pi t/4) + constant
if x = -4 when t = 0 then
-4 = -(4/pi) + c
c = (4/pi)-4
(I suspect you have another typo and the initial x was supposed to be 4/pi)
Hi! I need help with this problem:
For 0(smaller or equal to) t (smaller or equal to) 9, a particle moves along the x-axis. The velocity of the particle is given by v(t)= (sin pi/4 t). The particle is at position x= -4 when t=0.
a) For 0 (smaller or equal to) t (smaller or equal to) 0, when is the particle moving to the right?
Thanks!!
1 answer