Answered

The velocity of an object in meters per second varies directly with time in seconds since the object was dropped, as represented by the table.



\begin{table}

\captionsetup{labelformat=empty}

\caption{Velocity of a Falling Object}


Time (seconds)		Velocity (meters/second)

0		0

1		9.8

2		19.6

3		29.4

4		39.2




\end{table}



The acceleration due to gravity is the constant of variation. What is the acceleration due to gravity of a falling object?


A$4.9 \frac{m}{s^2}$
B$9.8 \frac{m}{s^2}$
C$10.2 \frac{m}{s^2}$
D$19.6 \frac{m}{s^2}$

Verified Answer :

  • Recognize that direct variation follows the formula \(v = k \cdot t\).
  • Select a data point, such as \(t=1, v=9.8\), and solve for \(k\): \(k = \frac{v}{t} = \frac{9.8}{1} = 9.8\).
  • Verify the result using another data point, \(t=4, v=39.2\), which confirms \(k = \frac{39.2}{4} = 9.8\).
  • Conclude that the acceleration due to gravity is \(\boxed{9.8 \text{ m/s}^2}\).

Explanation

  1. Understanding Direct Variation

The problem states that the velocity \(v\) of an object varies directly with the time \(t\) since it was dropped. This relationship can be expressed by the direct variation formula:
\[v = k \cdot t\]
where \(v\) is the velocity in meters per second, \(t\) is the time in seconds, and \(k\) is the constant of variation, which represents the acceleration due to gravity. We are given a table of data points relating time and velocity: \((0, 0), (1, 9.8), (2, 19.6), (3, 29.4),\) and \((4, 39.2)\). Our objective is to find the value of \(k\).

  1. Calculating the Constant of Variation

To determine the value of the constant of variation \(k\), we can select any non-zero data point from the table. Let us use the point where \(t = 1\) second and \(v = 9.8\) meters per second. Substituting these values into the direct variation formula:
\[9.8 = k \cdot 1\]
Solving for \(k\), we find:
\[k = \frac{9.8}{1} = 9.8\]
Thus, the acceleration due to gravity \(k\) is \(9.8 \text{ m/s}^2\).

  1. Verifying the Consistency

To ensure our result is consistent, we verify it with another data point from the table, such as \(t = 4\) seconds and \(v = 39.2\) meters per second:
\[39.2 = k \cdot 4\]
Dividing both sides by 4:
\[k = \frac{39.2}{4} = 9.8\]
The calculation consistently yields \(k = 9.8 \text{ m/s}^2\). This confirms that the relationship is indeed a direct variation with a constant acceleration due to gravity of \(9.8 \text{ m/s}^2\).

Examples

In physics, the constant acceleration due to gravity is a fundamental concept for understanding the motion of falling objects near Earth's surface. By observing how velocity increases linearly over time, we can determine this constant, which helps scientists and engineers predict the movement of everything from a falling apple to complex aerospace trajectories. This same mathematical approach of identifying constant rates of change can be applied to many other real-world scenarios, such as calculating the speed of a car moving at a constant acceleration or determining the cost per item in a bulk purchase.

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