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Chapter 2: Problem 21

A model rocket is launched straight upward. Its altitude \(y\) as a function of time is given by \(y=b t-c t^{2},\) where \(b=82 \mathrm{m} / \mathrm{s}\) \(c=4.9 \mathrm{m} / \mathrm{s}^{2}, t\) is the time in seconds, and \(y\) is in meters. (a) Use differentiation to find a general expression for the rocket's velocity as a function of time. (b) When is the velocity zero?

Short Answer

Expert verified
The velocity of the rocket is zero at \(t = 8.36734694\) seconds.

Step by step solution

01

Find the Velocity Function

Starting with the altitude-time function \(y = bt - ct^2\), differentiate it with respect to time \(t\) to obtain the velocity \(v\) as a function of time. The derivative of a linear term \(bt\) with respect to \(t\) is \(b\), and the derivative of a square term \(ct^2\) with respect to \(t\) is \(2ct\). Thus, the velocity function is given by \(v = b - 2ct\).
02

Set the Velocity Function to Zero

To find when the velocity is zero, set the velocity function to zero and solve for \(t\). That gives, \(0 = b - 2ct\), solving for \(t\) thus gives \(t = b / (2c)\). Substitute the given values \(b = 82 m/s\) and \(c = 4.9 m/s^2\) into the formula.
03

Solving for Time

Substitute \(b = 82 m/s\) and \(c = 4.9 m/s^2\) into \(t = b / (2c)\) to obtain \(t = 82 / (2*4.9)\). Solve the equation to find the value of \(t\).

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