Problem 84 - Fundamentals of Physics

Question

A rocket-driven sled running on a straight, levcl track is used to investigate the effects of large accelerations on humans. One such sled can attain a speed of \(1600 \mathrm{~km} / \mathrm{h}\) in \(1.8 \mathrm{~s}\), starting from rest. Find (a) the acceleration (assumed constant) in terms of \(g\) and (b) the distance traveled.

Step-by-Step Solution

Answer

Acceleration is approximately \(25.18g\) and the distance traveled is about 400 meters.

Step 1: Convert Speed to Meters per Second

First, convert the speed from kilometers per hour to meters per second. To do this, use the conversion: \[ 1 \, \text{km/h} = \frac{1000}{3600} \, \text{m/s}. \]Thus, the final speed is \[ 1600 \, \text{km/h} \times \frac{1000}{3600} \approx 444.44 \, \text{m/s}. \]

Step 2: Calculate Acceleration

We use the formula for acceleration \[ a = \frac{v_f - v_i}{t}, \]where \( v_f \) is the final velocity, \( v_i \) is the initial velocity (0 m/s since it starts from rest), and \( t \) is the time.Substitute the values into the formula:\[ a = \frac{444.44 \, \text{m/s} - 0 \, \text{m/s}}{1.8 \, \text{s}} \approx 246.91 \, \text{m/s}^2. \]

Step 3: Express Acceleration in Terms of g

Given that \( 1 \, g = 9.81 \, \text{m/s}^2 \), convert the acceleration to terms of \( g \):\[ a_g = \frac{246.91 \, \text{m/s}^2}{9.81 \, \text{m/s}^2} \approx 25.18g. \]Thus, the acceleration is approximately \( 25.18g \).

Step 4: Calculate Distance Traveled

For constant acceleration, use the formula \[ d = v_i \cdot t + \frac{1}{2} a t^2, \]where \( d \) is the distance. Since the initial velocity \( v_i = 0 \), the equation simplifies to:\[ d = \frac{1}{2} \times 246.91 \, \text{m/s}^2 \times (1.8 \, \text{s})^2. \]Calculate the distance:\[ d = \frac{1}{2} \times 246.91 \times 3.24 \approx 399.99 \, \text{m}. \]So, the distance traveled is approximately 400 meters.

Key Concepts

Acceleration

Acceleration is a key concept in kinematics, which describes how an object's velocity changes over time. When a rocket-driven sled starts from rest and reaches a high speed, it experiences acceleration. Acceleration can be calculated using the formula:

• \( a = \frac{v_f - v_i}{t} \)

This formula shows that acceleration \( a \) is the change in velocity \( v_f - v_i \) divided by the time \( t \) taken to change that velocity.
For the sled, starting from rest means the initial velocity \( v_i \) is 0. By using the final velocity \( v_f = 444.44 \text{ m/s} \) and time \( t = 1.8 \text{ s} \), we find the sled's acceleration to be \( 246.91 \text{ m/s}^2 \).
This high value signifies rapid acceleration, common in experiments studying the effects on humans.

Constant Acceleration

When we discuss constant acceleration, we refer to an unchanging rate of increase in velocity over time. Constant acceleration means the object speeds up at a steady rate. For the sled experiment, the assumption of constant acceleration simplifies calculations.
With constant acceleration, formulas become more predictable and reliable. For example, the distance an object travels can be accurately determined. In real-world scenarios, determining constant acceleration helps to model and predict motion more effectively. Understanding this concept is crucial for interpreting motion accurately in various scientific and engineering contexts.

Distance Traveled

To find out how far the sled traveled, we use the formula for distance under constant acceleration:

• \( d = v_i \cdot t + \frac{1}{2} a t^2 \)

In this scenario, with an initial velocity \( v_i = 0 \), the formula simplifies to \( d = \frac{1}{2} a t^2 \).
By substituting \( a = 246.91 \text{ m/s}^2 \) and \( t = 1.8 \text{ s} \) into the equation, we calculate the distance \( d \) to be approximately 400 meters.
Knowing how to calculate the distance traveled is crucial for predicting where an object will be after a given time, especially in safety tests like those with the sled.

Conversion of Units

Understanding unit conversion is fundamental in physics as it ensures that measurements and calculations are consistent and accurate. In the sled example, the initial speed of \( 1600 \text{ km/h} \) was converted to \( \text{m/s} \) using the conversion factor:

• \( 1 \text{ km/h} = \frac{1000}{3600} \text{ m/s} \)

This results in a speed of \( 444.44 \text{ m/s} \), which is a more practical unit for calculations in physics.
Converting units allows for a smoother transition into formulas that depend on the standard metric system, minimizing errors and confusion. Correct unit conversion helps ensure precision in experiments and calculations across scientific fields.