The model for the velocity of a shuttle, from liftoff at t = 0 s until the solid rocket boosters were jettisoned at t = 50.2 s, is given by
v(t)=0.001495833t^3−0.089395t^2+18.86t−0.52 (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 crank it out. The acceleration is
a(t) = 0.0044875t^2 - 0.17879t + 18.86
a(t) has a max/min when da/dt = 0
da/dt = 0.008975t - 0.17879
da/dt=0 at a=19.9209
a(10.9209) = 17.0792
Since a(t) is a parabola, there is no maximum value, but in the domain [0,50.2], the maximum value is
a(50.2) = 21.1934
To find the absolute maximum and minimum values of the acceleration, we need to consider the derivative of the velocity function. The derivative of a function represents the rate of change of that function.
Let's first find the derivative of the velocity function v(t). Taking the derivative of each term with respect to t:
v'(t) = 0.003491499t^2 - 0.17879t + 18.86
Now, v'(t) represents the acceleration function since it gives us the rate of change of velocity with respect to time. We can analyze this function to find the maximum and minimum values.
To find the critical points (where the acceleration could be maximum or minimum), we set v'(t) = 0 and solve for t:
0.003491499t^2 - 0.17879t + 18.86 = 0
This equation is a quadratic equation, which can be solved using the quadratic formula:
t = [-b ± √(b^2 - 4ac)] / (2a)
Here, a = 0.003491499, b = -0.17879, and c = 18.86.
Solving for t using the quadratic formula, we find two values of t:
t ≈ 28.72 s and t ≈ 84.59 s
Now, we need to check whether these critical points are within the range of our time interval (0 to 50.2 s).
Since 28.72 s falls within the range, we will consider it as a valid critical point. However, 84.59 s is beyond the given time interval, so we discard it from consideration.
To determine if the critical point at t = 28.72 s corresponds to a maximum or minimum value of acceleration, we can use the second derivative test. We need to find the second derivative of the velocity function:
v''(t) = 0.006982998t - 0.17879
Set v''(t) = 0 and solve for t:
0.006982998t - 0.17879 = 0
Simplifying, we find:
t ≈ 25.63 s
Since 25.63 s falls within the given time interval (0 to 50.2 s), we will consider it as a valid critical point.
Now, we can find the values of acceleration at the critical points.
Substitute t = 28.72 s into v'(t) to find the acceleration at this point:
v'(28.72) ≈ -0.2172 ft/s^2
Similarly, substitute t = 25.63 s into v'(t):
v'(25.63) ≈ 1.0153 ft/s^2
Hence, the absolute maximum value of acceleration occurs at approximately t = 25.63 s, with a value of about 1.0153 ft/s^2. The absolute minimum value of acceleration occurs at approximately t = 28.72 s, with a value of about -0.2172 ft/s^2.