A projectile is fired vertically upward and has a position given by s(t)=−16t^2+128t+192, for 0≤t≤9.


d. For what values of t on the interval [0, 9] is the instantaneous velocity positive (the projectile moves upward)?

Verified step by step guidance

1. insert step 1> Find the velocity function by differentiating the position function s(t) with respect to time t.
2. insert step 2> The position function is s(t) = -16t^2 + 128t + 192. Differentiate this to get the velocity function v(t).
3. insert step 3> The derivative of s(t) is v(t) = ds/dt = -32t + 128.
4. insert step 4> Set the velocity function v(t) > 0 to find when the projectile is moving upward.
5. insert step 5> Solve the inequality -32t + 128 > 0 to find the values of t for which the velocity is positive.

Key Concepts

Instantaneous Velocity

Instantaneous velocity is the rate of change of position with respect to time at a specific moment. It is calculated as the derivative of the position function, s(t). For the given function s(t) = -16t^2 + 128t + 192, finding the derivative s'(t) will provide the instantaneous velocity at any time t.

Derivative

The derivative of a function measures how the function's output changes as its input changes. In calculus, it is a fundamental tool for analyzing rates of change. For the position function s(t), the derivative s'(t) will yield a new function that describes the velocity of the projectile at any time t, allowing us to determine when the projectile is moving upward.

Critical Points and Intervals

Critical points occur where the derivative of a function is zero or undefined, indicating potential changes in the function's behavior. To find when the instantaneous velocity is positive, we need to analyze the sign of the derivative s'(t) over the interval [0, 9]. This involves identifying critical points and testing intervals to determine where the velocity is greater than zero.