A University of Rochester student decided to depart from Earth after his graduation to find work on Mars. Before building a shuttle, he conducted careful calculations. A model for the velocity of the shuttle, from liftoff at t = 0 s until the solid rocket boosters were jettisoned at t = 67.8929589270746 s, is given by
v(t)=0.00198833333333333t 3 −0.085875t 2 +26.23t−1.34 (in feet per second).
Using this model, estimate the absolute maximum value and absolute minimum value of the ACCELERATION of the shuttle between liftoff and the jettisoning of the boosters.
just plug and chug:
v=0.00198833333333333t^3 −0.085875t^2 +26.23t−1.34
since acceleration is v',
a = 0.005965t^2 - 0.17175t + 26.23
That's just a parabola opening up, with vertex at (14.3965,24.9937)
So, the minimum acceleration is 25, and the maximum will be at either t=0 or t=67...
I expect it will be the latter. Plug in that for t and evaluate v(t) there.
Well, buckle up because we're about to rocket into some calculations! Let's find the absolute maximum and minimum values of acceleration, shall we?
First, we need to find the derivative of the velocity function to get the acceleration function. So, let's differentiate v(t) with respect to t:
a(t) = 0.00198833333333333 * 3t^2 - 0.085875 * 2t + 26.23
Simplifying further:
a(t) = 0.00596499999999999t^2 - 0.17175t + 26.23
Now, to find the absolute maximum and minimum values of the acceleration, we need to find the critical points. These occur when the derivative of the acceleration function equals zero or is undefined.
Differentiating a(t) again:
a'(t) = 0.01193t - 0.17175
Setting a'(t) equal to zero:
0.01193t - 0.17175 = 0
0.01193t = 0.17175
t = 14.40 seconds
Now, we know that the rocket launches at t = 0 seconds, and the boosters are jettisoned at t = 67.8929589270746 seconds. So, the critical point is within this interval, and we need to check the endpoints too.
Now, let's evaluate the acceleration at these critical points:
a(14.40) = 0.00596499999999999 * (14.40)^2 - 0.17175 * 14.40 + 26.23
a(14.40) ≈ 25.5 ft/s^2
Now, let's evaluate the acceleration at the endpoints:
a(0) = 0.00596499999999999 * (0)^2 - 0.17175 * 0 + 26.23
a(0) ≈ 26.23 ft/s^2
a(67.8929589270746) = 0.00596499999999999 * (67.8929589270746)^2 - 0.17175 * 67.8929589270746 + 26.23
a(67.8929589270746) ≈ 25.9 ft/s^2
So, the absolute maximum acceleration value is approximately 26.23 ft/s^2, and the absolute minimum acceleration value is approximately 25.5 ft/s^2.
Remember, though, these are just estimates based on the given model. Actual conditions may vary, especially when dealing with rocket science. Good luck to the daring student, may their journey to Mars be filled with laughter and no clowning around! 🚀🤡
To find the absolute maximum and absolute minimum values of the acceleration of the shuttle, we need to find the critical points of the velocity function and evaluate the second derivative at those points.
The acceleration of an object is the derivative of its velocity function. Taking the derivative of the given velocity function:
v(t) = 0.00198833333333333t^3 - 0.085875t^2 + 26.23t - 1.34
a(t) = v'(t) = 0.00596499999999999t^2 - 0.17175t + 26.23
To find the critical points, we set the derivative equal to zero and solve for t:
0.00596499999999999t^2 - 0.17175t + 26.23 = 0
Using the quadratic formula, we find two potential critical points:
t = (-(-0.17175) ± √((-0.17175)^2 - 4(0.00596499999999999)(26.23))) / (2(0.00596499999999999))
Simplifying the equation:
t = (0.17175 ± √(0.17175^2 - 0.11989299))/ 0.01193
Calculating the discriminant:
0.17175^2 - 0.11989299 = 0.0294828766
Taking the square root:
√(0.0294828766) ≈ 0.171521
Calculating the critical points:
t₁ = (0.17175 - 0.171521) / 0.01193 ≈ 0.229278 seconds
t₂ = (0.17175 + 0.171521) / 0.01193 ≈ 28.7117 seconds
Now, we need to evaluate the second derivative of the velocity function, a(t), at these critical points to classify them as maximum or minimum:
a(t) = 0.00596499999999999t^2 - 0.17175t + 26.23
a''(t) = 2(0.00596499999999999)t - 0.17175
Evaluating a''(t₁):
a''(t₁) = 2(0.00596499999999999)(0.229278) - 0.17175 ≈ -0.114
Evaluating a''(t₂):
a''(t₂) = 2(0.00596499999999999)(28.7117) - 0.17175 ≈ 0.236
Since a''(t₁) is negative, it indicates a relative maximum point, and a''(t₂) is positive, indicating a relative minimum point.
Therefore, the relative maximum value of the acceleration is approximately 0.229278 seconds, and the relative minimum value of the acceleration is approximately 28.7117 seconds.
To find the absolute maximum and absolute minimum values of acceleration, we need to find the critical points of the function and determine whether they are maximum or minimum points.
The acceleration function can be found by taking the derivative of the velocity function with respect to time. Therefore, the acceleration function, a(t), is given by:
a(t) = v'(t) = 0.00198833333333333 * 3t^2 - 0.085875 * 2t + 26.23
To find the critical points, we need to find the values of t where a'(t) = 0. Taking the derivative of the acceleration function:
a'(t) = 0.00198833333333333 * 6t - 0.085875 * 2 = 0
Simplifying the equation:
0.011930 * t - 0.085875 * 2 = 0
0.011930 * t = 0.085875 * 2
t = (0.085875 * 2) / 0.011930
t ≈ 11.424 seconds
To determine whether this critical point is a maximum or minimum, we can analyze the behavior of the acceleration function before and after this point.
Before t = 11.424 seconds:
We can take the second derivative of the acceleration function to determine the concavity.
a''(t) = 0.011930 * 6 = 0.07158
Since the second derivative is positive, the function is concave up before the critical point.
After t = 11.424 seconds:
The second derivative doesn't change, so the function remains concave up after the critical point.
Therefore, the critical point is a minimum point of the acceleration function.
To find the absolute maximum and absolute minimum values of acceleration, we also need to evaluate the acceleration at the endpoints of the given interval.
At t = 0 seconds:
a(0) = 0.00198833333333333 * (0)^3 - 0.085875 * (0)^2 + 26.23
= -1.34 ft/s^2
At t = 67.8929589270746 seconds:
a(67.8929589270746) = 0.00198833333333333 * (67.8929589270746)^3 - 0.085875 * (67.8929589270746)^2 + 26.23
= 8.87 ft/s^2
Therefore, the absolute maximum value of acceleration is approximately 8.87 ft/s^2, occurring at t = 67.8929589270746 seconds, and the absolute minimum value of acceleration is -1.34 ft/s^2, occurring at t = 0 seconds.