At a basketball game, members of the spirit club use a t-shirt cannon to launch team t-shirts to spectators in the stands. the equation below describes the height, h, of the t-shirt, t seconds after it is shot from the cannon c feet above the ground, at an initial velocity of v. h = -16t 2 + vt + c solve the equation for initial velocity, v. the cannon launched a t-shirt from an initial height of 5 feet. after 3 seconds the shirt has reached a height of 131 feet. use your equation from #1 to find the initial velocity of the t-shirt. show your work. what does the equation from #1 tell you about the value of the velocity, v, when t = 0? explain.
The equation for initial velocity is s = ut + 1/2at^2, where s = distance, u = initial velocity, and a = acceleration.
Step-by-step explanation:
The two solutions are given as [tex]y_1(t)=\dfrac{1}{9}e^{-8t}+\dfrac{8}{9}e^{t}[/tex] and [tex]y_2(t)=\dfrac{-1}{9}e^{-8t}+\dfrac{1}{9}e^{t}[/tex]
Step-by-step explanation:
As the given equation is
[tex]y''+7y'-8y=0\\[/tex]
So the corresponding equation is given as
[tex]m^2+7m-8=0[/tex]
Solving this equation yields the value of m as
[tex](m+8)(m-1)=0\\m=-8, m=1[/tex]
Now the equation is given as
[tex]y(t)=C_1e^{m_1t}+C_2e^{m_2t}[/tex]
Here m1=-8, m2=1 so
[tex]y(t)=C_1e^{-8t}+C_2e^{t}[/tex]
The derivative is given as
[tex]y'(t)=-8C_1e^{-8t}+C_2e^{t}[/tex]
Now for the first case y(t=0)=1, y'(t=0)=0
[tex]y(t=0)=C_1e^{-8*0}+C_2e^{0}\\1=C_1+C_2\\\\y'(t=0)=-8C_1e^{-8*0}+C_2e^{0}\\0=-8C_1+C_2[/tex]
So the two equation of co-efficient are given as
[tex]C_1+C_2=1\\-8C_1+C_2=0[/tex]
Solving the equation yield
[tex]C_1=1/9 \\C_2=8/9[/tex]
So the function is given as
[tex]y_1(t)=\dfrac{1}{9}e^{-8t}+\dfrac{8}{9}e^{t}[/tex]
Now for the second case y(t=0)=0, y'(t=0)=1
[tex]y(t=0)=C_1e^{-8*0}+C_2e^{0}\\0=C_1+C_2\\\\y'(t=0)=-8C_1e^{-8*0}+C_2e^{0}\\1=-8C_1+C_2[/tex]
So the two equation of co-efficient are given as
[tex]C_1+C_2=0\\-8C_1+C_2=1[/tex]
Solving the equation yield
[tex]C_1=-1/9 \\C_2=1/9[/tex]
So the function is given as
[tex]y_2(t)=\dfrac{-1}{9}e^{-8t}+\dfrac{1}{9}e^{t}[/tex]
So the two solutions are given as [tex]y_1(t)=\dfrac{1}{9}e^{-8t}+\dfrac{8}{9}e^{t}[/tex] and [tex]y_2(t)=\dfrac{-1}{9}e^{-8t}+\dfrac{1}{9}e^{t}[/tex]
1. h = -16t² + vt + c
⇒ vt = 16t² + h - c
⇒ v = 16t + (h-c)/t
2. c = 5 ft
t = 3s
h = 131 ft
v = 16*3 + (131-5)/3 = 48 + 42 = 90 ft/s
when t=0, velocity is not defined. That means at t=0, there is no velocity of the t-shirt or it is 0.
a. [tex]P = 1000 - Ce^{-\frac{kt^2}{2} }[/tex]
b. [tex]C = 950[/tex]
c. [tex]P = 1000 - 950e^{-\frac{kt^2}{2} }[/tex]
Step-by-step explanation:
a. Let the number of penguins who have the disease t days after the outbreak be P
Initial number of penguins = 1000
Therefore, current number of penguins = 1000 - P
And the rate of spread of disease according to the statement is
[tex]\frac{dP}{dt}\alpha t(1000-P)\\\frac{dP}{dt}=kt(1000-P)[/tex]
where k is the constant of proportionality
[tex]\frac{dP}{1000-P}=kt.dt[/tex]
Integrating both sides
[tex]-ln(1000-P) = \frac{kt^2}{2}+c\\\frac{1}{(1000-P)} = Ce^{\frac{kt^2}{2} }\\ (1000-P) = Ce^{-\frac{kt^2}{2} }\\P = 1000 - Ce^{-\frac{kt^2}{2} }[/tex]
b. Seeing as 50 penguins had the disease initially,
t = 0
P = 50
The general solution of the differential solution becomes
50 = 1000 - C (anything raised to the power of 0 is 1, hence e is equal to 1)
[tex]C = 1000 - 50 = 950[/tex]
c. Therefore, the solution that satisfies the initial condition is
[tex]P = 1000 - 950e^{-\frac{kt^2}{2} }[/tex]
A) Judging by the slope field, it would appear the solution passing through (0, 2) would just be the line [tex]y=2[/tex]. This holds up for the given ODE, since if [tex]y(x)=2[/tex], then both sides of the ODE reduce to 0.
Since we can surmise that [tex]y=2[/tex] is an equilibrium for this ODE (that is, the derivative along this line is 0 everywhere), and the slope at (1, 0) is positive, we know that as [tex]x\to+\infty[/tex], the function [tex]y(x)[/tex] will converge to 2. In other words, as [tex]x[/tex] gets larger, the slope field suggests that the solution curve through (1, 0) will start to plateau and steadily approach [tex]y=2[/tex]. On the other hand, as [tex]x\to-\infty[/tex], the slope field tells us that the curve would rapidly diverge off to [tex]-\infty[/tex]. (When you actually draw the solution, you would end up with something resembling the plot of [tex]-e^{-x}[/tex].)
b) The tangent line to [tex]y=f(x)[/tex] at [tex]x=1[/tex], given that [tex]f(1)=0[/tex], takes the form
[tex]\ell(x)=f(1)+f'(1)(x-1)[/tex]
[tex]\ell(x)=f'(1)(x-1)[/tex]
When [tex]x=1[/tex], we have [tex]y=0[/tex], so
[tex]f'(1)=\dfrac13\cdot1\cdot(0-2)^2=\dfrac43[/tex]
and so the tangent line to [tex]f(x)[/tex] at [tex]x=1[/tex] is
[tex]\ell(x)=\dfrac43(x-1)[/tex]
Using the tangent line as an approximation, we would find
[tex]f(0.7)\approx\ell(0.7)=\dfrac43\left(\dfrac7{10}-1\right)=-\dfrac4{10}=-0.4[/tex]
c) The ODE is separable, so we can write
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac13x(y-2)^2\iff\dfrac{\mathrm dy}{(y-2)^2}=\dfrac x3\,\mathrm dx[/tex]
Integrating both sides gives us
[tex]-\dfrac1{y-2}=\dfrac{x^2}6+C[/tex]
Given that [tex]y(1)=0[/tex], we get
[tex]-\dfrac1{0-2}=\dfrac{1^2}6+C\implies C=\dfrac13[/tex]
so the particular solution is
[tex]-\dfrac1{y-2}=\dfrac{x^2}6+\dfrac13[/tex]
You're asked to find the solution in the form [tex]y=f(x)[/tex], so you should solve for [tex]y[/tex]. You would end up with
[tex]y=-\dfrac6{x^2+2}+2=\dfrac{2(x^2-1)}{x^2+2}[/tex]
h(t) = -16t^2 + vt + 5 h(3) = 131 ft h(3) = -16(3)^2 + 3v + 5 = 131 3v = 270 v = 90 ft/s
1.
[tex]h=-16t^2+vt+c\qquad\text{original equation}\\\\h+16t^2-c=vt\qquad\text{subtract terms not containing v}\\\\v=\dfrac{h+16t^2-c}{t}\qquad\text{divide by the coefficient of v}[/tex]
2. Filling in the given numbers, we have ...
[tex]v=\dfrac{131+16(3)^2-5}{3}=90\qquad\text{ft/s}[/tex]
3. The velocity is undefined when t=0. The equation is telling you it would take infinite velocity to travel some distance in zero time. The equation is meaningless in such a situation.
Initial velocity is 3.5. The equation is s = ut + 1/2at^2, where s - distance, u - inititial velocity, and a - acceleration. How do I modify the acceleration formula to one that gives me initial velocity? You can't change the acceleration formula to one that gives you the initial velocity you want, as a=v/t.
Step-by-step explanation:
h(t) = -16t^2 + vt + 5 h(3) = 131 ft h(3) = -16(3)^2 + 3v + 5 = 131 3v = 270 v = 90 ft/s
The required formula is [tex]v=\frac{h-c}{t}+16t[/tex]. The initial velocity is 90 ft/s. At t=0, the value of v is undefined.
Step-by-step explanation:
The equation below describes the height, h, of the t-shirt, t seconds after it is shot from the cannon c feet above the ground, at an initial velocity of v.
[tex]h=-16t^2+vt+c[/tex]
We need to solve this equation for v.
Subtract c from both sides.
[tex]h-c=-16t^2+vt[/tex]
Add 16t² on both sides.
[tex]h-c+16t^2=vt[/tex]
Divide both sides by t.
[tex]\frac{h-c+16t^2}{t}=v[/tex]
[tex]v=\frac{h-c}{t}+16t[/tex] .... (1)
Therefore the required formula is [tex]v=\frac{h-c}{t}+16t[/tex].
We need to find the velocity for
Initial height c = 5 feet
Time t = 3 sec
Height h = 131 feet
Substitute c = 5, t = 3 and h = 131 in equation (1).
[tex]v=\frac{131-5}{3}+16(3)[/tex]
[tex]v=42+48=90[/tex]
Therefore the initial velocity is 90 ft/s.
Substitute t=0 in equation (1).
[tex]v=\frac{h-c}{0}+16(0)[/tex]
[tex]v=\infty[/tex]
At t=0, the value of v is undefined.