Problem 86 - Fundamentals of Physics

Question

A motorcyclist who is moving along an \(x\) axis directed toward the east has an acceleration given by \(a=(6.1-1.2 t) \mathrm{m} / \mathrm{s}^{2}\) for \(0 \leq t \leq 6.0 \mathrm{~s}\). At \(t=0,\) the velocity and position of the cyclist are \(2.7 \mathrm{~m} / \mathrm{s}\) and \(7.3 \mathrm{~m}\). (a) What is the maximum speed achieved by the cyclist? (b) What total distance does the cyclist travel between \(t=0\) and \(6.0 \mathrm{~s} ?\)

Step-by-Step Solution

Answer

(a) The maximum speed is approximately 18 m/s. (b) The total distance traveled is about 48.1 m.

Step 1: Find the Velocity Function

To find the velocity as a function of time, we need to integrate the acceleration function, \( a(t) = 6.1 - 1.2t \), with respect to time. The velocity function is given by \( v(t) = \int (6.1 - 1.2t) \, dt = 6.1t - 0.6t^2 + C \). We calculate the constant \( C \) using the initial condition \( v(0) = 2.7 \, \mathrm{m/s} \):\( 2.7 = 6.1(0) - 0.6(0)^2 + C \Rightarrow C = 2.7 \). Therefore, \( v(t) = 6.1t - 0.6t^2 + 2.7 \).

Step 2: Determine Maximum Speed

The maximum speed occurs at the peak of the velocity function. To find when this happens, we set the derivative of \( v(t) \) with respect to \( t \) to zero: \( \frac{dv}{dt} = 6.1 - 1.2t = 0 \). Solving for \( t \), we get \( t = \frac{6.1}{1.2} \approx 5.08 \, \mathrm{s} \). Substitute this back into the velocity function: \( v(5.08) = 6.1(5.08) - 0.6(5.08)^2 + 2.7 \approx 18 \, \mathrm{m/s} \).

Step 3: Set Up Position Function

Now, we find the position function by integrating the velocity function \( v(t) = 6.1t - 0.6t^2 + 2.7 \). Integrating, \( x(t) = \int (6.1t - 0.6t^2 + 2.7) \, dt = 3.05t^2 - 0.2t^3 + 2.7t + C \). Use the initial condition \( x(0) = 7.3 \, \mathrm{m} \) to find \( C \): \( 7.3 = 3.05(0)^2 - 0.2(0)^3 + 2.7(0) + C \Rightarrow C = 7.3 \). Thus, \( x(t) = 3.05t^2 - 0.2t^3 + 2.7t + 7.3 \).

Step 4: Calculate Total Distance Traveled

To calculate the total distance traveled by the cyclist from \( t = 0 \) to \( t = 6 \), evaluate \( x(t) \) at these times: \( x(6) = 3.05(6)^2 - 0.2(6)^3 + 2.7(6) + 7.3 \). Simplifying, \( x(6) \approx 55.4 \, \mathrm{m} \). The initial position is \( x(0) = 7.3 \, \mathrm{m} \), thus the total distance traveled is \( 55.4 - 7.3 = 48.1 \, \mathrm{m} \).

Key Concepts

Acceleration Function

In kinematics, the acceleration function is crucial to understanding how a body's speed changes over time. It describes how quickly an object is speeding up or slowing down. For this exercise, the acceleration function is given by \[ a(t) = 6.1 - 1.2t \] where \( a(t) \) is the acceleration at time \( t \). This function shows that the motorcyclist starts at a positive acceleration and decreases over time, which implies that the motorcyclist's speed will initially increase and then decrease as time goes on.

• Positive values indicate speeding up in a positive direction.
• The linear decrease suggests a consistent slowing of acceleration.

Understanding how the acceleration changes helps in determining how the velocity behaves over time, as acceleration directly influences the rate of change of velocity.

Velocity Function

The velocity function represents the object's speed at any given moment and its change across time. To determine the velocity function, we integrate the acceleration function with respect to time: \[ v(t) = \ \int (6.1 - 1.2t) \, dt = 6.1t - 0.6t^2 + C \]Here, \( C \) is a constant that we find using initial conditions. A known initial velocity or position gives us the exact velocity function, represented as: \[ v(t) = 6.1t - 0.6t^2 + 2.7 \] after solving for \( C \) using the initial velocity \( v(0) = 2.7 \, \mathrm{m/s} \).
This integration process and the resulting function are fundamental in determining how far an object will travel over a specific period, which is particularly essential if the acceleration is not constant.

Integration in Calculus

Integration is a critical concept in calculus for solving kinematics problems. It allows us to find one function from its derivative. In the context of this exercise, we use integration to derive the velocity function from the acceleration function and the position function from the velocity function.

• The indefinite integral of the acceleration yields the velocity function.
• The indefinite integral of the velocity provides the position function.

These integrals often include an arbitrary constant \( C \), which is determined using known initial conditions (position or velocity at a specific time). Not only does integration fill the gap between acceleration and position but it also enables us to calculate the distance traveled and the area under the curve of the function outlining these changes.

Maximum Speed Calculation

To determine the maximum speed achieved, we need to look at the velocity function's behavior over time. This often involves finding where the slope of the velocity-time graph is zero, indicating a peak. Mathematically, the peak point corresponds to the time \( t \) at which the derivative of the velocity function is zero: \[ \frac{dv}{dt} = 6.1 - 1.2t = 0 \]Solving gives \( t = \frac{6.1}{1.2} \approx 5.08 \, \mathrm{s} \). This tells us the time when the velocity function reaches its maximum point. Substituting this \( t \) value back into the velocity function gives the maximum speed: \[ v(5.08) = 6.1(5.08) - 0.6(5.08)^2 + 2.7 \approx 18 \, \mathrm{m/s} \]At this point, the acceleration ceases to positively contribute to the velocity, resulting in a peak speed. Understanding these calculations helps predict the behavior of objects in motion, providing valuable insight for various applications in physics and engineering.