Solving Problem with Vectors - Velocity & Accel.

[ncr7]

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?

 

[D H]

You implicitly assumed $t<=56/45$
when you wrote
$$\sqrt{(5.6t-4.5t^2)^2} = 5.6t-4.5t^2$$
What happens if $t>56/45$?

 

[ncr7]

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?

 

[D H]

My hint wasn't good enough.

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

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