Question

The equation of motion for a float in a parade is \(x=-9.2 \mathrm{~m}+(1.5 \mathrm{~m} / \mathrm{s}) t\). (a) Where is the float at \(t=3.5 \mathrm{~s}\) ? (b) At what time is the float at \(x=0\) ?

Short answer

At \(t = 3.5\) s, the float is at \(-3.95\) m; it reaches\(x = 0\) m at \(t \approx 6.13\) s.

Step-by-step solution

Identify the given equation and interpret

The equation given is the equation of motion for the float: \[x = -9.2 \, \text{m} + (1.5 \, \text{m/s}) t\]Here, "\( x \)" represents the position of the float as a function of time "\( t \)", measured in meters and seconds, respectively. We will use this equation to find the float's position at a specific time and the time it reaches a position of zero meters.

Calculate float's position at specified time

To find where the float is at \(t = 3.5 \, \mathrm{s}\), substitute \(t = 3.5\) into the equation:\[x = -9.2 + (1.5 \cdot 3.5)\]Calculate:\[x = -9.2 + 5.25 = -3.95 \, \mathrm{m}\]Thus, at \(t = 3.5 \, \mathrm{s}\), the float is at \(-3.95 \, \mathrm{m}\).

Determine the time when float is at x = 0

To find when the float is at \(x = 0\), set the equation to zero and solve for \(t\):\[0 = -9.2 + 1.5t\]Add 9.2 to both sides:\[9.2 = 1.5t\]Divide both sides by 1.5:\[t = \frac{9.2}{1.5} \approx 6.13 \, \mathrm{s}\]Thus, the float reaches \(x = 0\) at approximately \(t = 6.13 \, \mathrm{s}\).

Key concepts

Equation of Motion

The equation of motion is a fundamental tool in kinematics. It helps describe how an object's position changes over time. In our exercise, the equation for the float is:
\[x = -9.2 \text{ m} + (1.5 \text{ m/s}) t\]
This simple linear equation tells us that the float starts at an initial position of \(-9.2\) meters.
The presence of the "\(+ (1.5 \text{ m/s}) t\)" part indicates a uniform motion, with a constant speed of \(1.5\) meters per second in the positive direction.
Using these types of equations allows us to predict an object's future or past positions based on given conditions.

• Initial Position: The value where the object starts, often represented by a constant term, like \(-9.2\) in this case.
• Rate of Change: Represented by the coefficient of \(t\), indicating the speed at which position changes over time.

Time Calculation

Calculating time in kinematics involves determining when an object reaches a certain position. From our exercise, the expression \(x= 0\) means we are solving for the time when the float is exactly at the start line or reference point.
By setting the motion equation to zero, we reverse-engineer the formula to isolate \(t\):
\[0 = -9.2 + 1.5t\]
Adding 9.2 to both sides gives us the distance traveled due to motion, which yields:
\[9.2 = 1.5t\]
Finally, dividing both sides by 1.5 will solve for time:

• Calculate: \[ t = \frac{9.2}{1.5} \approx 6.13 \, \text{s}\]

This means the float reaches the starting position or point \(x=0\) at \(6.13\) seconds. Understanding this process involves recognizing the relationship between position, time, and constant speed in linear motion.

Displacement Calculation

Displacement refers to an object's change in position. As used in the exercise, displacement calculation means finding out where the float is at a specific time. For instance, determining the float's position at \(t=3.5\) seconds involves substituting the time in the equation of motion.
For the float, the formula is substituted as follows:

• Substitute time in equation: \[x = -9.2 + (1.5 \times 3.5)\]
• Simplify the calculation: \[x = -9.2 + 5.25 = -3.95 \, \text{m}\]

This tells us that at \(3.5\) seconds, the float is at \(-3.95\) meters.
This information not only gives the float's current position but also the relative position to its start, effectively showcasing the float's displacement.