Question

The acceleration of a bus is given by \(a_{x}(t)=\alpha t\) where \(\alpha=1.2 \mathrm{~m} / \mathrm{s}^{3} .\) (a) If the bus's velocity at time \(t=1.0 \mathrm{~s}\) is \(5.0 \mathrm{~m} / \mathrm{s},\) what is its velocity at time \(t=2.0 \mathrm{~s} ?\) (b) If the bus's position at time \(t=1.0 \mathrm{~s}\) is \(6.0 \mathrm{~m},\) what is its position at time \(t=2.0 \mathrm{~s} ?\) (c) Sketch \(a_{y}-t, v_{y}-t,\) and \(x-t\) graphs for the motion.

Short answer

(a) 6.8 m/s at t = 2.0 s; (b) 11.8 m at t = 2.0 s; (c) Quadratic and cubic curves for velocity and position.

Step-by-step solution

Understand the given acceleration function

We are given that the acceleration function is \(a_x(t) = \alpha t\), where \(\alpha = 1.2 \, \mathrm{m/s^3}\). This tells us how the acceleration varies with time. We need to integrate this function to find velocity and position as functions of time.

Integrate acceleration to find velocity

The velocity function \(v_x(t)\) can be found by integrating the acceleration function. Integrate \(a_x(t) = 1.2t\) with respect to \(t\). The integral is: \[v_x(t) = \int 1.2t \, dt = \frac{1.2}{2}t^2 + C = 0.6t^2 + C\]where \(C\) is the constant of integration determined by initial conditions.

Use initial velocity to find constant C

At \(t = 1.0 \, \mathrm{s}\), the velocity \(v_x(1) = 5.0 \, \mathrm{m/s}\). Substitute these values into the velocity equation:\[5.0 = 0.6(1.0)^2 + C\]Solve for \(C\): \[5.0 = 0.6 + C \rightarrow C = 4.4\]Thus, \(v_x(t) = 0.6t^2 + 4.4\).

Calculate velocity at t = 2.0 s

Substitute \(t = 2.0\) into the velocity equation:\[v_x(2.0) = 0.6(2.0)^2 + 4.4 = 2.4 + 4.4 = 6.8 \, \mathrm{m/s}\]Thus, the velocity at \(t = 2.0 \, \mathrm{s}\) is \(6.8 \, \mathrm{m/s}\).

Integrate velocity to find position

To find the position function \(x(t)\), integrate the velocity function \(v_x(t) = 0.6t^2 + 4.4\):\[x(t) = \int (0.6t^2 + 4.4) \, dt = 0.2t^3 + 4.4t + C_2\]where \(C_2\) is another constant of integration.

Use initial position to find constant C2

At \(t = 1.0 \, \mathrm{s}\), the position \(x(1) = 6.0 \, \mathrm{m}\). Substitute these values into the position equation:\[6.0 = 0.2(1.0)^3 + 4.4(1.0) + C_2\]Solve for \(C_2\):\[6.0 = 0.2 + 4.4 + C_2 \rightarrow C_2 = 1.4\]Thus, \(x(t) = 0.2t^3 + 4.4t + 1.4\).

Calculate position at t = 2.0 s

Substitute \(t = 2.0\) into the position equation:\[x(2.0) = 0.2(2.0)^3 + 4.4(2.0) + 1.4 = 1.6 + 8.8 + 1.4 = 11.8 \, \mathrm{m}\]Thus, the position at \(t = 2.0 \, \mathrm{s}\) is \(11.8 \, \mathrm{m}\).

Sketch the a-t, v-t, and x-t graphs

The graph of \(a_x(t) = 1.2t\) is a straight line through the origin with slope 1.2. The graph of \(v_x(t) = 0.6t^2 + 4.4\) is a quadratic curve opening upwards. The graph of \(x(t) = 0.2t^3 + 4.4t + 1.4\) is a cubic curve. Plot these functions with respective variable on the y-axis and time \(t\) on the x-axis.

Key concepts

Acceleration

Acceleration is a measure of how quickly an object's velocity changes with time. In kinematics, acceleration is often provided as a function of time. In this exercise, the acceleration is given by the formula \( a_x(t) = \alpha t \), where \( \alpha = 1.2 \; \mathrm{m/s^3} \).
This means the acceleration increases linearly over time, making it a time-dependent acceleration.
This linear relationship is represented as a straight line on an acceleration-time graph, starting from the origin and having a slope of 1.2.

• The slope indicates how steeply acceleration increases over time.
• Positive slope means acceleration increases with time.

Understanding acceleration is crucial as it lays the foundation for calculating velocity and position through integration.

Velocity

Velocity is the rate at which an object's position changes over time and is itself dependent on acceleration. From acceleration \( a(t) = \alpha t \), we can find velocity by integrating the acceleration function.
Performing this integration gives us the velocity function: \( v_x(t) = 0.6t^2 + C \). The constant of integration \( C \) is determined using initial conditions, like the initial velocity provided.

• The velocity at \( t = 1.0 \; \mathrm{s} \) is 5.0 m/s, which helps us solve for \( C \).
• Once found, it allows us to determine the velocity at any other time, such as at \( t = 2.0 \; \mathrm{s} \).

By substituting into the velocity equation, we find \( v_x(2.0) = 6.8 \; \mathrm{m/s} \). Integrating acceleration to find velocity demonstrates how changes in motion evolve over time.

Position

Position describes the location of an object over time. With kinematics, you can determine position by integrating the velocity function.
From our earlier step, the velocity function is \( v_x(t) = 0.6t^2 + 4.4 \). Integrating the velocity function yields the position function: \( x(t) = 0.2t^3 + 4.4t + C_2 \).

• The constant \( C_2 \) is found using initial position conditions.
• Given that \( x(1.0) = 6.0 \; \mathrm{m} \), we calculate \( C_2 = 1.4 \).

This gives us the full position function \( x(t) = 0.2t^3 + 4.4t + 1.4 \).
Calculating at \( t=2.0 \; \mathrm{s} \), we find the position to be 11.8 m.
Position, therefore, accumulates the effect of velocity over time, just as velocity accumulates the effect of acceleration.

Integration

Integration in kinematics is the process used to derive velocity from acceleration and position from velocity. It works by summing up the infinitesimal changes over time.
The integral of the acceleration function \( a(t) \) gives you the velocity function, and similarly, integrating the velocity function provides the position function.

• Integration introduces a constant of integration, \( C \), which is essential to adjust according to initial conditions, making each solution unique to its scenario.
• Each integration step transforms the linear, quadratic, or cubic relationships shown in the kinematics equations.

This integral process allows us to track the motion from acceleration to velocity and then to position, illustrating how motion equations are interconnected.

Graphs of Motion

Graphs of motion visually depict relationships between acceleration, velocity, and position over time, enhancing our understanding of kinematic equations.
Each type of graph provides unique insights.

• The acceleration-time graph is a straight line, indicating uniform increase over time.
• The velocity-time graph is a parabola opening upward reflecting increasing velocity as a function of time squared.
• The position-time graph is cubic, showing how position accumulates complexly over time.

Visualizing these graphs can help clarify the motion characteristics and the effect of acceleration and initial conditions on velocity and position.