# Physics

Under what conditions can a quadratic function be a perfect square?

Also..
A stone is thrown vertically upward at a speed of 40m/s. How long does it take to reach a height of 90m?

I ended up with a quadratic of t^2-8t+18 which cannot be factored...So is the answer that the stone doesn't reach a height of 90m?

Another question...
A hockey player shoots a puck from his own end of the ice with an initial speed of 10 m/s. The ice surface is 60 m long and produces a deceleration of 1.0m/s^2 on the sliding puck.
a) Will the puck travel the length of the ice?
b) With what speed will it reach the far end of the ice OR how far from the end of the ice will it come to rest?

I think I used the wrong equation.
I used V2^2 = V1^2 + 2ad
And solved for distance to get 50 m which indicates that the puck didn't travel the length of the ice. But I have a feeling I did it wrong since I'm also supposed to use a kinematic equation for the second question rather than just subtraction.

Thanks!

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1. I do not know what perfect square means
I guess you are using 10 instead of 9.8 for g to get round numbers. That is ok

h = 90 = 0 + 40 t -5 t^2

5 t^2 - 40 t + 90 = 0

t^2 - 8 t + 18 = 0 check

solve by completing the square or quadratic equation. I will complete the square

t^2 - 8 t = -18
t^2 - 8 t + 4^2 = -18+16
(t-4)^2 = -2
t = 4 +/- sqrt (-2)
Yes, sqrt(-2) is imaginary number. Stone does not get that high.

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posted by Damon
2. A hockey player shoots a puck from his own end of the ice with an initial speed of 10 m/s. The ice surface is 60 m long and produces a deceleration of 1.0m/s^2 on the sliding puck.
a) Will the puck travel the length of the ice?
b) With what speed will it reach the far end of the ice OR how far from the end of the ice will it come to rest?
++++++++++++++++++++++===
How far will it go?
v = Vo - a t = 10 - 1 t
0 = 10 - t when it stops, so it stops after 10 seconds
d = Vo t - (1/2)(1) t^2
d = 10 (10) - .5 (100)
d = 100 - 50 = 50 meters to stop.
It does not make 60 meters.
It stops after 50 meters

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posted by Damon
3. Thanks!

Can you check if I'm doing this question right and help me complete it. I'm having difficulty with the last question.

Two cars are approaching one another in the same lane of a highway. One has a speed of 30m/s, the other a speed of 20m/s when they are 96m apart. The drivers simultaneously apply brakes to avert a collision. Both cars lose speed at 6.0m/s^2. How long after the application of brakes do the cars collide? Where are the cars and at what speeds are they travelling when they collide?

dA = v1t + 1/2 at^2
= 30t - 3t^2

dB = v1t + 1/2 t^2
= 20t - 3t^2

dA+dB = 96
30t - 3t^2 + 20t - 3t^2 = 96
50t - 6t^2 = 96
- 6t^2 + 50t - 96 = 0
*dividing both sides by -2*
3t^2 - 25t + 48
Should I factor with the quadratic formula?

And then I have no idea how to answer this question -
Where are the cars and at what speeds are they travelling when they collide?

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