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!
To determine the conditions under which a quadratic function can be a perfect square, we need to understand the concept of perfect square trinomials. A perfect square trinomial is a quadratic expression that can be factored into two identical binomials. In general, a quadratic function in the form of ax^2 + bx + c can be a perfect square if and only if its discriminant (i.e., b^2 - 4ac) is zero.
In the case of the quadratic function t^2 - 8t + 18, the discriminant is calculated as follows:
Discriminant = (-8)^2 - 4(1)(18) = 64 - 72 = -8
Since the discriminant is negative, this quadratic cannot be factored into identical binomials, indicating that it is not a perfect square trinomial.
Now, let's move on to the second question about the stone thrown vertically upward. To determine the time it takes for the stone to reach a height of 90m, we can use the equation of motion for free-falling bodies:
h = v0t + (1/2)gt^2
Here, h represents the final height (90m), v0 is the initial velocity of the stone (40m/s), g is the acceleration due to gravity (-9.8m/s^2), and t is the time we want to find.
Substituting the given values into the equation, we get:
90 = 40t - (1/2)(9.8)t^2
Rearranging the equation in the standard quadratic form:
0 = (1/2)(9.8)t^2 - 40t + 90
To solve for t, we can either factor the quadratic or use the quadratic formula. If you were not able to factor it, let's use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = (1/2)(9.8) = 4.9, b = -40, and c = 90.
Plugging the values into the formula:
t = (-(-40) ± √((-40)^2 - 4(4.9)(90))) / (2(4.9))
t = (40 ± √(1600 - 1764)) / 9.8
t = (40 ± √(-164)) / 9.8
Since the value under the square root is negative, the resulting values of t will be complex numbers. Therefore, the stone does not reach a height of 90m.
Moving on to the third question about the hockey player shooting a puck. To determine if the puck will travel the length of the ice, we need to find the distance it will cover. Since the ice surface produces a deceleration of 1.0m/s^2, the equation of motion we should use is:
v^2 = u^2 + 2as
Here, v is the final velocity, u is the initial velocity (10m/s), a is the acceleration (-1.0m/s^2), and s is the distance we want to find (60m).
Plugging the values into the equation, we get:
0 = (10)^2 + 2(-1.0)s
0 = 100 - 2s
2s = 100
s = 50
As you correctly calculated, the puck will not travel the full length of the ice and will come to rest 50m from the end of the ice.
To summarize:
1. The quadratic function t^2 - 8t + 18 is not a perfect square trinomial.
2. The stone does not reach a height of 90m.
3. The puck does not travel the full length of the ice and comes to rest 50m from the end of the ice.