A weight is attached to an elastic spring that is suspended from a ceiling. If the weight s pulled 1 inch below its rest position and released its displacement in inches after t seconds is given by x(t)=(5pt+p/3. Find the first two times for which the displacement s 1.5 inches.
fix the typo and we can proceed from there. You can use π to copy/paste for pi.
to Steve A weight is attached to an elastic spring that is suspended from a ceiling. If the weight is pulled 1 inch below its rest position and released its displacement in inches after t seconds is given by x(t)= 2cos(5ðt+ð/3). Find the first two times for which the displacement is 1.5 inches.
to Steve A weight is attached to an elastic spring that is suspended from a ceiling. If the weight is pulled 1 inch below its rest position and released its displacement in inches after t seconds is given by x(t)= 2cos(5ðt+ð/3). Find the first two times for which the displacement is 1.5 inches.
I cannot write the pi sign and when I try copying and paste it it would change into that theta looking sign between 5 and t
To find the first two times for which the displacement is 1.5 inches, we need to solve the equation x(t) = 1.5.
Given that the displacement function is x(t) = (5p)t + (p/3), we can substitute this expression into the equation:
(5p)t + (p/3) = 1.5
Now, let's solve this equation for the variable t.
First, we'll simplify the equation:
5pt + p/3 = 1.5
Next, let's isolate the term with t by subtracting p/3 from both sides:
5pt = 1.5 - p/3
Now, let's divide both sides by 5p:
t = (1.5 - p/3) / (5p)
This equation gives us the value of t when the displacement is 1.5 inches. However, we want to find the first two times when the displacement is 1.5 inches. Therefore, we need to find two distinct values of t that satisfy the equation.
To find the first value of t, we can substitute p = 1 into the equation:
t = (1.5 - 1/3) / (5 * 1)
Simplifying further, we get:
t = (1.5 - 1/3) / 5
t = (4.5 - 1/3) / 15
t = (4/3) / 15
t = 4 / (3 * 15)
t = 4 / 45
t = 1 / 11.25
t ≈ 0.089
So, the first time when the displacement is 1.5 inches is approximately 0.089 seconds.
To find the second value of t, we can substitute p = 2 into the equation:
t = (1.5 - 2/3) / (5 * 2)
Simplifying further, we get:
t = (1.5 - 2/3) / 10
t = (4.5 - 2/3) / 30
t = (12/3) / 30
t = 4/30
t = 2/15
t ≈ 0.133
So, the second time when the displacement is 1.5 inches is approximately 0.133 seconds.