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!!
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)
To determine when the particle is moving to the right, we need to find the values of t for which the velocity function, v(t), is positive.
Given that v(t) = sin(π/4t), we want to find the values of t where v(t) > 0.
Using trigonometric properties, we know that sin(x) > 0 when x lies in the interval (0, π).
So, we can set π/4t > 0 and solve for t.
π/4t > 0
π/4 > 0 (since t ≠ 0)
Since the angle π/4 is already positive, we don't need to flip the inequality symbol. We can write the solution as:
0 < t < ∞
Therefore, the particle is moving to the right for all values of t greater than 0.