Problem 15 - Technical Mathematics with Calculus

Question

For these problems, we use Eq. \(1018, s=v_{0} t+a t^{2} / 2\) which says that, if assume the initial velocity is zero, the distance \(s\) fallen by a body (from rest) varies directly as the square of the elapsed time \(t\) If a body falls \(176 \mathrm{m}\) in \(6.00 \mathrm{s}\), how far will it fall in \(9.00 \mathrm{s} ?\)

Step-by-Step Solution

Answer

The body will fall 396.09 m in 9.00 s.

Step 1: Identify the known values

The problem states that the body falls a distance of 176 m in 6.00 s with an initial velocity of zero. We can denote the initial velocity as \(v_0 = 0\), the time as \(t_1 = 6.00\) s, and the distance fallen as \(s_1 = 176\) m. We need to find the distance \(s_2\) when \(t_2 = 9.00\) s.

Step 2: Use the given equation to find the acceleration

Substitute the known values \(v_0 = 0\), \(t_1 = 6.00\) s, and \(s_1 = 176\) m into Eq. 1018: \(s_1 = v_0 t_1 + \frac{1}{2} a t_1^2\). Since \(v_0 = 0\), the equation simplifies to \(s_1 = \frac{1}{2} a t_1^2\). We can solve for acceleration \(a\) by rearranging the equation as \(a = \frac{2 s_1}{t_1^2}\).

Step 3: Calculate the acceleration

Substitute the known values into the rearranged equation to find \(a\): \(a = \frac{2 * 176}{6.00^2} = \frac{352}{36} \approx 9.78 \mathrm{m/s^2}\).

Step 4: Use the acceleration to find the distance fallen in 9.00 s

Now that we have the acceleration, we can use Eq. 1018 again to find the distance \(s_2\) fallen in \(t_2 = 9.00\) s. Substituting \(v_0 = 0\), \(t_2 = 9.00\) s, and the calculated \(a\) into Eq. 1018: \(s_2 = v_0 t_2 + \frac{1}{2} a t_2^2\), it simplifies to \(s_2 = \frac{1}{2} a t_2^2\).

Step 5: Calculate the distance fallen in 9.00 s

Finally, plug the known values of \(a\) and \(t_2\) into the simplified equation to find \(s_2\): \(s_2 = \frac{1}{2} * 9.78 * 9.00^2 = \frac{1}{2} * 9.78 * 81 = 396.09 \mathrm{m}\). The body will fall 396.09 m in 9.00 s.

Key Concepts

Understanding Free Fall Motion

Free fall motion is a key concept in kinematics, describing the motion of objects falling solely under the influence of gravity. Without air resistance, every object in free fall near the Earth's surface accelerates downwards at the same rate, regardless of its mass.

In our example, an object falls 176 m in 6 seconds, showcasing that free fall is uniformly accelerated motion, which means that velocity increases linearly over time while the distance covered is proportional to the square of the time elapsed. This relationship is described by the kinematic equation for uniformly accelerated motion, which simplifies in free fall with an initial velocity of zero to: \( s = \frac{1}{2} a t^2 \).

To help students better grasp free fall motion, it's important to emphasize the following points:

• Objects in free fall experience constant acceleration.
• The impact of air resistance is typically ignored in basic problems.
• The initial velocity in free fall problems is often zero when the object starts from rest.
• Direction is important; since the object is falling, we typically take 'downward' as positive when defining our coordinates.

Understanding these principles allows students to tackle free fall problems confidently, applying the kinematic equation to determine the distance fallen over a specified time interval.

Acceleration Due to Gravity

The acceleration due to gravity, generally represented as \( g \), is the constant acceleration that acts on objects when they are in free fall near the Earth's surface. On Earth, this value is approximately \( 9.81 \, \text{m/s}^2 \) and is sometimes rounded to \( 9.8 \, \text{m/s}^2 \) for simplicity in calculations.

In the exercise example, we calculated the acceleration as \( 9.78 \, \text{m/s}^2 \), which is very close to the standard \( g \) value. Small discrepancies can arise due to rounding or measurement differences. For educational purposes, highlighting the following points is essential:

• Gravity is a universal force acting on all mass.
• \( g \) is approximately consistent near the Earth's surface, but it varies slightly with altitude and latitude.
• All objects experience the same gravitational acceleration, ignoring air resistance.

When students grasp how gravity consistently accelerates objects in free fall, they can apply this understanding to predict motion in a variety of contexts.

Quadratic Equations in Kinematics

Quadratic equations are often encountered in kinematics when dealing with accelerated motion, as acceleration results in the squared dependence of distance on time. These equations are in the general form \( ax^2 + bx + c = 0 \) and play an essential role in predicting the motion of objects.

In the context of our falling body problem, we use a simplified version of the quadratic equation where the initial velocity is zero, reducing the equation to \( s = \frac{1}{2} a t^2 \), with \( s \) being the distance fallen, \( a \) the constant acceleration, and \( t \) the time.

To help students handle quadratic equations in kinematics:

• Emphasize the importance of identifying known and unknown quantities.
• Ensure students understand how to rearrange equations to solve for the desired variable.
• Illustrate the process of substituting known values to solve for unknowns.
• Discuss how quadratic equations reflect the properties of accelerated motion, like the curvature of an object's path or its increasing speed over time.

Understanding quadratic equations enables students to solve a wide range of motion problems beyond just free fall scenarios.