Problem 153 - Calculus Volume 1

Question

A rocket is fired vertically upward from the ground. The distance s in feet that the rocket travels from the ground after \(t\) seconds is given by \(s(t)=-16 t^{2}+560 t\) . a. Find the velocity of the rocket 3 seconds after being fired. b. Find the acceleration of the rocket 3 seconds after being fired.

Step-by-Step Solution

Answer

Velocity at 3s: 464 ft/s; Acceleration at 3s: -32 ft/s².

Step 1: Differentiate the position function to find the velocity function

The position function is given by \(s(t) = -16t^2 + 560t\). To find the velocity function \(v(t)\), we need to differentiate \(s(t)\) with respect to \(t\). This gives:\[v(t) = \frac{ds}{dt} = -32t + 560\].

Step 2: Calculate the velocity at t = 3 seconds

Now that we have the velocity function, \(v(t) = -32t + 560\), substitute \(t = 3\) to find the velocity of the rocket 3 seconds after being fired:\[v(3) = -32(3) + 560 = -96 + 560 = 464\, \text{ft/s}\].

Step 3: Differentiate the velocity function to find the acceleration function

The velocity function is \(v(t) = -32t + 560\). To find the acceleration function \(a(t)\), we differentiate \(v(t)\) with respect to \(t\). This gives:\[a(t) = \frac{dv}{dt} = -32\].

Step 4: Calculate the acceleration at t = 3 seconds

Since the acceleration \(a(t)\) is constant and does not depend on \(t\), \(a(t) = -32\, \text{ft/s}^2\) for all values of \(t\). Thus, the acceleration of the rocket 3 seconds after being fired is \(-32\, \text{ft/s}^2\).

Key Concepts

Differentiation

Differentiation is a key concept in calculus used to determine the rate at which a function is changing at any given point. Essentially, it allows us to find the derivative of a function, which gives us valuable information about the function's behavior, such as velocity from a position function.

When differentiating, we focus on how one quantity changes relative to another. For example, in our rocket problem, the position function, given by \(s(t) = -16t^2 + 560t\), represents how the rocket's height changes with time. By differentiating this function with respect to time \(t\), we obtain the velocity function, revealing the speed and direction of the rocket at any moment.

• To differentiate, apply rules like the power rule to each term of the function.
• The derivative of \(-16t^2\) is \(-32t\) and the derivative of \(560t\) is \(560\).

The result of this differentiation process is the velocity function: \(v(t) = -32t + 560\).

Velocity

Velocity is a vector quantity that describes the rate of change of the position of an object. Unlike speed, which is scalar, velocity includes both magnitude and direction.

In the context of the rocket's problem, the velocity function \(v(t) = -32t + 560\) tells us how fast the rocket is moving upwards or downwards at any given second. By substituting specific values of \(t\) into this function, we can calculate the rocket's speed at different times.

• At \(t = 3\) seconds, the velocity is calculated as follows:
• Substitute 3 into the velocity function: \(v(3) = -32(3) + 560 = 464\) ft/s.
• The positive value indicates an upward movement at that moment.

Thus, the rocket's velocity 3 seconds after being fired is 464 ft/s upwards.

Acceleration

Acceleration is the rate at which an object's velocity changes with time, and it can be constant or variable. Differentiating the velocity function gives us the acceleration function, reflecting how velocity changes.

For the rocket, once we have the velocity function \(v(t) = -32t + 560\), we differentiate it to find the acceleration function \(a(t)\).

• The derivative of the constant velocity coefficient \(-32t\) is \(-32\).
• This results in a constant acceleration: \(a(t) = -32\) ft/s².

The negative sign indicates the rocket is slowing down (deceleration) due to gravity. Notably, since \(a(t)\) is constant, the acceleration remains \(-32\) ft/s² for any time \(t\), including at 3 seconds.

Position Function

The position function represents the height or position of an object over time. In problems like our rocket scenario, the position function is critical in determining how far an object travels over time.

Given as \(s(t) = -16t^2 + 560t\), this quadratic function tells us where the rocket is at any moment after it launches. The coefficients reflect the effects of gravity and initial velocity.

• The \(-16t^2\) term is attributed to gravitational acceleration, accounting for the downward force.
• The \(560t\) term reflects the initial upward velocity of the rocket.

By analyzing this position function, we can predict not just the velocity at certain points, but the overall path of the rocket as it ascends, slows, and eventually descends.