A bug is placed on the road. t seconds after it is put down, the function is p(t) = -16tsquared + 64t, where p(t) is measure in feet. At what moment will the bug be 14 feet from where it began?
p(t) same as f(x) basically just put in 14 for 7 in the equation but don't put it in for the t in p(t) solve out and that is your answer
p(t)=-16(14)^2 + 64(14)
oh whoops i read the question wrong scratcch that
oh okay here it is sorry about the above;
p(t)= - 16t^2 +64t
14= - 16t^2 + 64t bring 14 to other side and factor or use the equation -b = or minus square root of (b - 4AC)and divide by 2A and you should get what t is
14= - 16t^2 + 64t
16 t^2 - 64 T + 14 = 0
8 t^2 -32 t + 7 = 0
t = [32 +/-sqrt (1024-224)]/16
= [32 +/- sqrt(800)]/16
= [32 +/- 20sqrt 2]/16
= [ 32 +/- 28.3]/16
= 3.76 on the way down
and
= .231 on the way up
To find the moment when the bug is 14 feet from where it began, we need to find the value of t when p(t) = 14.
Given the function p(t) = -16t^2 + 64t, we can set it equal to 14 and solve for t:
-16t^2 + 64t = 14
First, let's rearrange the equation to put it in standard quadratic form:
-16t^2 + 64t - 14 = 0
Next, we can solve this quadratic equation using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a
Here, a = -16, b = 64, and c = -14. Substituting these values into the formula, we get:
t = (-64 ± √(64^2 - 4*(-16)*(-14))) / (2*(-16))
Simplifying further:
t = (-64 ± √(4096 - 896)) / (-32)
t = (-64 ± √3200) / (-32)
Now, let's calculate the square root of 3200:
√3200 ≈ 56.57
Finally, substituting this value:
t = (-64 ± 56.57) / (-32)
t_1 = (-64 + 56.57) / (-32) ≈ 0.23
t_2 = (-64 - 56.57) / (-32) ≈ 3.18
Therefore, the bug will be 14 feet from where it began at approximately 0.23 seconds and 3.18 seconds after it is put down.