Problem with vectors

  • Thread starter ncr7
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  • #1
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The velocity of a particle moving in the xy plane is given by = (5.6 t - 4.5 t2) + 8.2, with in meters per second and t (> 0) in seconds. (a) What is the acceleration when t = 3.3 s? (b) When (if ever) is the acceleration zero? (c) When (if ever) does the speed equal 10 m/s?

I got the first 2 but when I went to find c I can't seem to get it.

I took

10m/s = sqrt((5.6t - 4.5t^2)^2 + (8.2)^2
10^2 = (sqrt((5.6t - 4.5t^2)^2 + (8.2)^2)^2
100 = (5.6t - 4.5t^2)^2 + 67.24)
32.76 = (5.6t - 4.5t^2)^2
sqrt(32.76) = sqrt((5.6t - 4.5t^2)^2)
5.72 = 5.6t - 4.5t^2
0 = -4.5t^2 + 5.6t - 5.72
then I used quadratic to find out the two values...
but I get a negative in the sqrt on the quadratic and I can't do that... any suggestions?
 

Answers and Replies

  • #2
D H
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You implicitly assumed [itex]t<=56/45[/itex]
when you wrote
[tex]\sqrt{(5.6t-4.5t^2)^2} = 5.6t-4.5t^2[/tex]
What happens if [itex]t>56/45[/itex]?
 
  • #3
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so how do you figure it out? I'm not sure how you would go about doing it.

Could you take

5.6t-4.5t^2
t(5.6-4.5t)
t = 5.6/4.5 ?

if so what happens with the 5.72 on the other side of the equation?
 
  • #4
D H
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My hint wasn't good enough.

It is not valid to write [itex]\sqrt{(5.6t-4.5t^2)^2} = 5.6t-4.5t^2[/itex] when [itex]t>56/45[/itex] because [itex]5.6t-4.5t^2[/itex] is negative.

For example, try [itex]t=2[/itex]. The velocity vector at [itex]t=2[/itex] is [itex][ 5.6*2 - 4.5*2^2, 8.2 ] = [ -6.8, 8.2 ] [/itex]. Note that the [itex]x[/itex] component of the velocity is negative. [itex]\sqrt{(-6.8)^2} = 6.8,\text{\ not\ } -6.8[/itex].
 
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