Question

1. the altitude (in meters) of a rocket t sec into flight is given by

$s(t) = -2t^3 + 10t^2 + 400t + 80, t \geq 0$

a. find an expression $v(t)$ for the rocket’s velocity at any time t.
b. compute the rocket’s velocity when $t = 5$.
c. find the maximum altitude attained by the rocket.

Explanation:

Step1: Find the velocity function

The velocity \(v(t)\) is the derivative of the position function \(s(t)\). Using the power rule \((x^n)^\prime = nx^{n - 1}\), for \(s(t)=-2t^{3}+10t^{2}+400t + 80\), we have:
\(v(t)=s^\prime(t)=-2\times3t^{2}+10\times2t + 400\times1+0\)
\(v(t)=-6t^{2}+20t + 400\)

Step2: Compute the velocity at \(t = 5\)

Substitute \(t = 5\) into \(v(t)\):
\(v(5)=-6\times5^{2}+20\times5 + 400\)
\(v(5)=-6\times25+100 + 400\)
\(v(5)=-150+100 + 400\)
\(v(5)=350\)

Step3: Find the maximum altitude

First, find the critical points of \(s(t)\) by setting \(v(t)=0\), i.e., \(-6t^{2}+20t + 400 = 0\). Divide through by \(-2\) to get \(3t^{2}-10t - 200=0\).
Using the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(a = 3\), \(b=-10\), \(c=-200\):
\(t=\frac{10\pm\sqrt{(-10)^{2}-4\times3\times(-200)}}{2\times3}=\frac{10\pm\sqrt{100 + 2400}}{6}=\frac{10\pm\sqrt{2500}}{6}=\frac{10\pm50}{6}\)
We get \(t=\frac{10 + 50}{6}=10\) or \(t=\frac{10-50}{6}=-\frac{20}{3}\). Since \(t\geq0\), we consider \(t = 10\).
Now, find \(s(10)\):
\(s(10)=-2\times10^{3}+10\times10^{2}+400\times10 + 80\)
\(s(10)=-2000+1000+4000 + 80\)
\(s(10)=3080\)

Answer:

a. \(v(t)=-6t^{2}+20t + 400\)
b. \(v(5) = 350\)
c. The maximum altitude is \(3080\) meters.