91Ó°ÊÓ

Acceleration is the rate at which an object's velocity changes over time. Imagine you're on a bus that starts from rest and gradually speeds up. The way this speed increases is governed by the acceleration. For this exercise, the bus's acceleration depends on time, given by the formula \( a_x(t) = 1.2t \; \mathrm{m/s^3} \). This equation tells us that as time goes on, the acceleration increases linearly.

- **Time-dependent Acceleration**: This means the longer the time, the higher the acceleration. A steeper slope on an acceleration-time graph indicates a larger acceleration value.

In practical terms, if you were in the bus, you'd feel it pushing you back in your seat more forcefully as time goes by. The formula gives us insight into how quickly and strongly the bus's speed is changing from moment to moment.

Velocity is the speed of the bus with a specific direction. To find out how fast the bus is going at a given time, we need to use its acceleration. By integrating the acceleration function, we determine the velocity function \( v(t) \). Integration involves finding how acceleration builds up over time to affect the speed. For this problem:

- **Integrating Acceleration**: The integration of \( a_x(t) = 1.2t \) gives us \( v(t) = 0.6t^2 + C \), where \( C \) is a constant determined by initial conditions.

- **Finding the Constant**: Given the velocity is \( 5.0 \; \mathrm{m/s} \) at \( t = 1.0 \; \mathrm{s} \), we determine that \( C = 4.4 \). Hence, the velocity function becomes \( v(t) = 0.6t^2 + 4.4 \).

This tells us that the velocity starts at 5.0 \( \mathrm{m/s} \) at \( t = 1.0 \; \mathrm{s} \) and then increases as a function of time, becoming 6.8 \( \mathrm{m/s} \) at \( t = 2.0 \; \mathrm{s} \). The parabolic shape of the velocity-time graph shows this upward trend.

Position tells us where the bus is at a specific time on the road. To get the position, we integrate the velocity function, just as we integrated acceleration to get velocity. The position function \( x(t) \) results from this integration process, representing how the bus's location changes over time. In this case:

- **Integration of Velocity**: Integrating \( v(t) = 0.6t^2 + 4.4 \), gives \( x(t) = 0.2t^3 + 4.4t + D \), with \( D \) as a constant based on initial conditions.

- **Setting Initial Conditions**: When given that the initial position at \( t = 1.0 \; \mathrm{s} \) is \( 6.0 \; \mathrm{m} \), we find \( D = 1.4 \). Therefore, the function becomes \( x(t) = 0.2t^3 + 4.4t + 1.4 \).

This position function shows that as time increases, the position of the bus doesn't just follow a straight path but instead a cubic curve. At \( t = 2.0 \; \mathrm{s} \), the bus is at 11.8 \( \mathrm{m} \) from the starting point.