Question

Velocity functions A projectile is fired vertically upward into the air, and its position (in feet) abous the ground after \(t\) seconds is given by the function \(s(t)\) a. For the following functions \(s(t)\). find the instantaneous velocity function \(v(t) .\) (Recall that the velocity function \(v\) is the derivative of the position function \(s\) ) Determine the instantancous velocity of the projectile at \(t=1\) and \(t=2\) seconds. $$s(t)=-16 t^{2}+100 t$$

Short answer

Answer: The instantaneous velocities at \(t=1\) and \(t=2\) are \(68\text{ ft/s}\) and \(36\text{ ft/s}\), respectively.

Step-by-step solution

Find the derivative of the position function

To find the instantaneous velocity function \(v(t)\), we need to find the derivative of the position function \(s(t)\). Differentiate \(s(t)\) with respect to \(t\). $$ v(t) = \frac{ds(t)}{dt} $$

Calculate the derivative of the given position function

Now, differentiate the given position function \(s(t)=-16t^2+100t\) with respect to \(t\): $$ v(t) = \frac{d}{dt}(-16t^2 + 100t) $$ Using the power rule, we have: $$ v(t) = -32t + 100 $$ This is the instantaneous velocity function \(v(t)\).

Calculate the instantaneous velocity at \(t=1\) and \(t=2\)

Now, we need to evaluate the instantaneous velocity function \(v(t)\) at \(t=1\) and \(t=2\). Plug in \(t=1\) and \(t=2\) into \(v(t)\): At \(t=1\): $$ v(1) = -32(1) + 100 = -32 + 100 = 68\text{ ft/s} $$ At \(t=2\): $$ v(2) = -32(2) + 100 = -64 + 100 = 36\text{ ft/s} $$

Final results

The instantaneous velocity function is \(v(t) = -32t + 100\). The instantaneous velocity of the projectile is \(68\text{ ft/s}\) at \(t=1\) and \(36\text{ ft/s}\) at \(t=2\).

Key concepts

Derivative

The derivative is a fundamental component in calculus. It essentially provides us with information about how a function changes. To understand derivatives, think of them as a way to calculate the slope of a curve. In simpler terms:

• It tells you the rate at which one quantity changes in respect to another.
• For a position function, this rate of change is particularly useful in determining velocity.

When we find the derivative of a position function, like in our exercise, we're figuring out how the position changes over time. It's like having a magic wand that reveals all speed changes of a moving object.

Velocity Function

The velocity function is directly linked to the derivative of a position function. Simply put, if a position function tells us where an object is at any time, the velocity function tells us how fast it is moving and in which direction.

• The function represents the instantaneous rate of change of the object's position.
• In mathematical language, it's the derivative of the position function with respect to time.

For the given exercise, we derived the velocity function by differentiating the position function: \[v(t) = \frac{d}{dt}(-16t^2 + 100t) = -32t + 100\]This simplified function can now be used to evaluate the speed of the projectile at any specific time.

Instantaneous Velocity

Instantaneous velocity is all about the precise speed of an object at an exact moment in time. Imagine you have a snapshot of a car at a single point on its journey – the instantaneous velocity tells you exactly how fast it's going right at that point.

• This concept is different from average velocity, which considers the entire journey.
• It's found by evaluating the velocity function at a specific time.

In our case, the instantaneous velocities at times \(t=1\) and \(t=2\) were calculated as:- At \(t=1\): \[v(1) = -32(1) + 100 = 68 \text{ ft/s}\]- At \(t=2\): \[v(2) = -32(2) + 100 = 36 \text{ ft/s}\]These calculations show exactly how fast the projectile was moving at those precise times. As time changes, so does the instantaneous velocity, highlighting the dynamic nature of motion.