Respuesta :
Answer:
a. The rock's velocity is [tex]v(t)=36-1.6t \:{(m/s)}[/tex] and the acceleration is [tex]a(t)=-1.6 \:{(m/s^2)}[/tex]
b. It takes 22.5 seconds to reach the highest point.
c. The rock goes up to 405 m.
d. It reach half its maximum height when time is 6.59 s or 38.41 s.
e. The rock is aloft for 45 seconds.
Step-by-step explanation:
- Velocity is defined as the rate of change of position or the rate of displacement. [tex]v(t)=\frac{ds}{dt}[/tex]
- Acceleration is defined as the rate of change of velocity. [tex]a(t)=\frac{dv}{dt}[/tex]
a.
The rock's velocity is the derivative of the height function [tex]s(t) = 36t - 0.8 t^2[/tex]
[tex]v(t)=\frac{d}{dt}(36t - 0.8 t^2) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\v(t)=\frac{d}{dt}\left(36t\right)-\frac{d}{dt}\left(0.8t^2\right)\\\\v(t)=36-1.6t[/tex]
The rock's acceleration is the derivative of the velocity function [tex]v(t)=36-1.6t[/tex]
[tex]a(t)=\frac{d}{dt}(36-1.6t)\\\\a(t)=-1.6[/tex]
b. The rock will reach its highest point when the velocity becomes zero.
[tex]v(t)=36-1.6t=0\\36\cdot \:10-1.6t\cdot \:10=0\cdot \:10\\360-16t=0\\360-16t-360=0-360\\-16t=-360\\t=\frac{45}{2}=22.5[/tex]
It takes 22.5 seconds to reach the highest point.
c. The rock reach its highest point when t = 22.5 s
Thus
[tex]s(22.5) = 36(22.5) - 0.8 (22.5)^2\\s(22.5) =405[/tex]
So the rock goes up to 405 m.
d. The maximum height is 405 m. So the half of its maximum height = [tex] \frac{405}{2} =202.5 \:m[/tex]
To find the time it reach half its maximum height, we need to solve
[tex]36t - 0.8 t^2=202.5\\36t\cdot \:10-0.8t^2\cdot \:10=202.5\cdot \:10\\360t-8t^2=2025\\360t-8t^2-2025=2025-2025\\-8t^2+360t-2025=0[/tex]
For a quadratic equation of the form [tex]ax^2+bx+c=0[/tex] the solutions are
[tex]x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
[tex]\mathrm{For\:}\quad a=-8,\:b=360,\:c=-2025:\\\\t=\frac{-360+\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2-\sqrt{2}\right)}{4}\approx 6.59\\\\t=\frac{-360-\sqrt{360^2-4\left(-8\right)\left(-2025\right)}}{2\left(-8\right)}=\frac{45\left(2+\sqrt{2}\right)}{4}\approx 38.41[/tex]
It reach half its maximum height when time is 6.59 s or 38.41 s.
e. It is aloft until s(t) = 0 again
[tex]36t - 0.8 t^2=0\\\\\mathrm{Factor\:}36t-0.8t^2\rightarrow -t\left(0.8t-36\right)\\\\\mathrm{The\:solutions\:to\:the\:quadratic\:equation\:are:}\\\\t=0,\:t=45[/tex]
The rock is aloft for 45 seconds.
