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How do I find the point where the direction of the particle changes? In desperate need of help.?
From what I have set up so far:
For 0<x<6, a particle is moving along the x-axis with the velocity v(t)=2sin(e^(t/4)+1 and a(t)=1/2e^(t/4)cos(e^t/4). (I worked these out, but they are definitely right.)
From that follows that s(t)=2cos(e^(t/4) * e(t/4)/4 +t
The average velocity between (0,6) was asked. I used the velocity formula and then inserted for
(f(b)-f(a))/(b-a) and received the right solution, which is 0.0202292. No units are given.
Now for the ones I'm not sure about.
Next was asked how far is the distance travelled in the time interval. I just integrated v(t) (which is how I got s(t)), for the interval (0,6) and received the amount 1.73409.
For the one I am stumped on:
The particle changes direction exactly once within the time interval. I assumed this would just be at v(t)=0, and even plotted the graph. According to the graph, it should occur at or around t=5.2
My calculations, on the other hand, go as follows:
0= 2sin(e^(t/4))+1
-0.5=sin(e^(t/4)
asin(-0.5) =-30 =e(t/4)
which does not make sense because for no value of t will e ever be negative because it will only become smaller approacing 0.
Next I tried findign positions on the unit circle for which it could work with. So far no luck, and I have a feeling I am overcomplicating things. Please have mercy on my soul an help me out? :D
1 Antwort
- EMLv 7vor 7 JahrenBeste Antwort
I'll take it from this step:
-0.5 = sin[e^(t/4)]
Arcsin only gets you one solution; in this case, it's one that doesn't make sense. You need an expression to represent all possible solutions and ultimately one such that 0 < t < 6. And the angle must be in radians.
-π/6 + 2πn = e^(t/4) or -5π/6 + 2πn = e^(t/4)
t = 4 ln(2πn - π/6) or t = 4 ln(2πn - 5π/6)
n = 0 gives the nonsensical solution. Trying n = 1,
t = 4 ln(2π - π/6) or t = 4 ln(2π - 5π/6)
t = 4 ln(11π/6) or t = 4 ln(7π/6)
t = 7.00 or t = 5.20
t = 5.20 just as your graph shows.