Problem 78 - AP Calculus with 8 Practice Tests 15th

Question

Find the acceleration (in \(\mathrm{ft} / \mathrm{sec}^{2}\) ) needed to bring a particle moving with a velocity of \(75 \mathrm{ft} / \mathrm{sec}\) to a stop in 5 seconds. (A) -3 (B) -6 (C) -15 (D) -25

Step-by-Step Solution

Answer

The necessary acceleration is \(-15 \, \mathrm{ft/s^2}\), which is option (C).

Step 1: Understand the Problem

We are given the initial velocity of a particle as \( 75 \, \mathrm{ft/s} \) and the final velocity needs to be \( 0 \, \mathrm{ft/s} \) after \( 5 \, \mathrm{seconds} \). We need to find the acceleration that causes this change in velocity.

Step 2: Use the Formula for Acceleration

The formula to calculate acceleration \( a \) when you know the initial velocity \( v_i \), final velocity \( v_f \), and time \( t \) is:\[ a = \frac{v_f - v_i}{t} \]

Step 3: Substitute the Given Values into the Formula

Substitute \( v_f = 0 \, \mathrm{ft/s} \), \( v_i = 75 \, \mathrm{ft/s} \), and \( t = 5 \, \mathrm{s} \) into the formula:\[ a = \frac{0 - 75}{5} \]

Step 4: Calculate the Acceleration

Perform the arithmetic operation:\[ a = \frac{-75}{5} = -15 \, \mathrm{ft/s^2} \]

Step 5: Verify if the Acceleration Meets the Criteria

The result of \(-15 \, \mathrm{ft/s^2}\) meets the requirement that the particle comes to a stop in the given time. It also matches option (C) from the list.

Key Concepts

Initial Velocity

Initial velocity, often represented as \( v_i \), is the speed at which an object is moving when it begins an observation. In our exercise, the initial velocity is given as \( 75 \, \text{ft/s} \). It marks how fast our particle is moving forward before any forces, like deceleration, are applied to change its state of motion.

This concept is crucial because it serves as the launching pad for our calculations. Knowing where you start gives you the basis for figuring out how much change is needed to get to a desired state, like stopping. It's similar to knowing your starting point on a journey, so you can plan the path ahead.

• Initial velocity is denoted by \( v_i \).
• In this exercise, \( v_i = 75 \, \text{ft/s} \).
• This value sets the stage for understanding the changes in motion.

Final Velocity

Final velocity, signified as \( v_f \), is the speed at which an object moves at the end of the observed period. For our particle, the final velocity must be \( 0 \, \text{ft/s} \) because it comes to a stop.

Reaching a final velocity of zero means the particle has halted its movement completely. This is essential in calculations because to find acceleration, you need to know both the initial and final velocities. Think of it as knowing your destination's exact point when traveling; it tells you precisely when to stop.

• Final velocity is denoted by \( v_f \).
• For this problem, the particle's \( v_f = 0 \, \text{ft/s} \).
• It helps in determining the amount of change required in motion.

Time

Time, used in these calculations as \( t \), is the duration over which the change in velocities occurs. The exercise specifies this as \( 5 \, \text{seconds} \). Time is a key factor in evaluating acceleration because it measures how quickly the velocity changes.

Understanding time in the concept of motion tells us how long the change is taking place. In real-life terms, if your change in speed should happen faster, you need to alter your acceleration accordingly. Without knowing the time factor, you can't accurately compute acceleration.

• Time is denoted by \( t \).
• Here, \( t = 5 \, \text{seconds} \).
• It determines the pace of change in velocity.