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!
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.
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?
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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
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?
To determine under what conditions a quadratic function can be a perfect square, we need to analyze its factors. A quadratic function can be written in the form:
f(x) = (ax+b)^2
Expanding this equation gives us:
f(x) = a^2x^2 + 2abx + b^2
For the quadratic function to be a perfect square, the coefficient of x^2 (a^2) should be a perfect square and the constant term (b^2) should also be a perfect square. This means that the quadratic function can be a perfect square if both a and b are perfect squares.
Now let's move on to the stone's upward motion problem. To find the time it takes for the stone to reach a height of 90m, we can use the kinematic equation:
h = V0t - (1/2)gt^2
Where:
h is the height (90m),
V0 is the initial velocity (40m/s),
g is the acceleration due to gravity (-9.8m/s^2),
and t is the time.
Substituting the known values into the equation, we get:
90 = (40)t - (1/2)(-9.8)t^2
Rearranging the equation, we have:
(1/2)(-9.8)t^2 - (40)t + 90 = 0
Simplifying and multiplying by 2 to eliminate the fraction:
-9.8t^2 + 80t - 180 = 0
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
Plugging in the values a = -9.8, b = 80, and c = -180, we can calculate the time it takes for the stone to reach a height of 90m.
Now, going back to your quadratic equation t^2 - 8t + 18, if it cannot be factored, then it indeed doesn't have real roots. Therefore, the stone does not reach a height of 90m.
Moving on to the hockey player's puck problem, you correctly used the initial equation V2^2 = V1^2 + 2ad. However, this equation is applicable only for uniform acceleration. In this case, the deceleration of the puck is not uniform, so we need to use a different equation.
The appropriate kinematic equation for this scenario is the following:
V2 = V1 + at
Where:
V2 is the final velocity of the puck,
V1 is the initial velocity of the puck (10m/s),
a is the acceleration of the puck (-1.0m/s^2), and
t is the time taken for the puck to come to rest.
Since we want to find the final velocity of the puck when it reaches the far end of the ice, we set V2 equal to zero. Now we can solve for t:
0 = 10 - 1.0t
Rearranging the equation:
t = 10 / 1.0
t = 10 seconds
Since the time taken for the puck to come to rest is 10 seconds and the ice surface is 60m long, we can calculate the distance traveled:
Distance = V1 * t + (1/2) * a * t^2
Distance = 10 * 10 + (1/2) * (-1.0) * 10^2
Distance = 100 - 50
Distance = 50m
Therefore, the puck will come to rest 50m from the end of the ice.
I hope this explanation helps you understand the solutions to these problems! If you have any further questions, please let me know.