Problem 55 - College Physics

Question

The rocket driven sled Sonic Wind No. 2, used for investigating the physiological effects of large accelerations, runs on a straight, level track that is \(1080 \mathrm{~m}\) long. Starting from rest, it can reach a speed of \(1610 \mathrm{~km} / \mathrm{h}\) in \(1.80 \mathrm{~s}\). (a) Compute the acceleration in \(\mathrm{m} / \mathrm{s}^{2}\) and in \(g\) 's. (b) What is the distance covered in 1.80 s? (c) A magazine article states that, at the end of a certain run, the speed of the sled decreased from \(1020 \mathrm{~km} / \mathrm{h}\) to zero in \(1.40 \mathrm{~s}\) and that, during this time, its passenger was subjected to more than \(40 g .\) Are these figures consistent?

Step-by-Step Solution

Answer

(a) Acceleration is 248.46 m/s² or 25.33 g's. (b) Distance is 400.64 m. (c) Deceleration is only 20.63 g's.

Step 1: Convert Speed to m/s

Convert the initial speed of the sled from km/h to m/s. To do this, use \( v = 1610 \text{ km/h} \times \frac{1000}{3600} \), which simplifies to \( v \approx 447.22 \text{ m/s} \). For the deceleration part, the initial speed \( 1020 \text{ km/h} \) converts to \( 283.33 \text{ m/s} \).

Step 2: Calculate Acceleration

Use the formula for acceleration \( a = \frac{v - u}{t} \) where \( v \) is the final velocity, \( u \) is the initial velocity, and \( t \) is time. For the acceleration, with \( u = 0 \), \( v = 447.22 \text{ m/s} \), and \( t = 1.80 \text{ s} \), \( a = \frac{447.22}{1.80} \approx 248.46 \text{ m/s}^2 \).

Step 3: Express Acceleration in g's

To convert acceleration to g's, divide by the acceleration due to gravity \( g \approx 9.81 \text{ m/s}^2 \). So, \( 248.46 \text{ m/s}^2 \approx 25.33 \text{ g} \).

Step 4: Calculate Distance Covered

Use the formula \( s = ut + \frac{1}{2}at^2 \). With \( u = 0 \), \( a = 248.46 \text{ m/s}^2 \), and \( t = 1.80 \text{ s} \), the distance is \( s = 0 + \frac{1}{2} \times 248.46 \times (1.80)^2 \approx 400.64 \text{ m} \).

Step 5: Verify Deceleration Figures

For deceleration, use \( a = \frac{0 - 283.33}{1.40} \approx -202.38 \text{ m/s}^2 \). Convert \(-202.38 \text{ m/s}^2\) to g's: \(-202.38 / 9.81 \approx -20.63 \text{ g} \), which is not more than 40 g as stated.

Key Concepts

Acceleration Calculation

Understanding acceleration involves determining how quickly an object is increasing its speed. Mathematically, acceleration is calculated using the formula \[ a = \frac{v - u}{t} \] where:

• \( v \) is the final velocity
• \( u \) is the initial velocity
• \( t \) is the time over which the change occurs

The result from this formula will give you acceleration in meters per second squared (m/s²).
In our exercise, the sled started from rest, so the initial speed \( u \) was 0, and increased to \( 447.22 \text{ m/s} \) over \( 1.80 \text{ s} \). This yielded an acceleration of about \( 248.46 \text{ m/s}^2 \).
To express acceleration in terms of gravitational force, or \( g \)'s, you divide the calculated acceleration by the standard acceleration due to gravity \( 9.81 \text{ m/s}^2 \), resulting in approximately \( 25.33 \text{ g} \).
This method helps to relate the rocket sled’s acceleration to the familiar force of gravity.

Kinematics Equations

Kinematics is a branch of mechanics that describes the motion of points, bodies, and systems of bodies, without considering the forces involved.
Key equations help solve problems related to motion, such as determining acceleration or the distance traveled by an object.
One important equation used in our exercise is the distance formula: \[ s = ut + \frac{1}{2}at^2 \]
Here:

• \( s \) is the distance covered
• \( u \) is the initial velocity
• \( a \) is the acceleration
• \( t \) is the time taken

With the sled having an initial velocity of zero, acceleration of \( 248.46 \text{ m/s}^2 \), and a time of \( 1.80 \text{ s} \), the distance was calculated as \( 400.64 \text{ m} \).
These equations form the foundation of understanding how objects move under constant acceleration and are fundamental in physics problem solving scenarios.

Unit Conversion

Unit conversion is an essential skill in physics. It ensures that all physical quantities are expressed in compatible units, which is crucial when performing calculations.
In physics problems, speeds are often converted from kilometers per hour (km/h) to meters per second (m/s) because the SI unit for speed is m/s.
The conversion process applies this relation: \[ 1 \text{ km/h} = \frac{1000}{3600} \text{ m/s} \]
For the sled, the initial speed of \( 1610 \text{ km/h} \) was converted to \( 447.22 \text{ m/s} \), making it easier to compute other quantities using standard equations.
Likewise, the initial deceleration speed of \( 1020 \text{ km/h} \) was converted to \( 283.33 \text{ m/s} \). Learning these conversions helps streamline problem-solving processes and avoid errors when interpreting data.

Deceleration Analysis

Deceleration is just negative acceleration, meaning an object is slowing down. Analyzing deceleration involves the same principles as calculating acceleration, with attention to sign changes since deceleration is negative.
Using the formula: \[ a = \frac{v - u}{t} \]with \( v = 0 \), it helped determine how quickly the sled came to a stop.
In the given scenario, the sled decreased from \( 283.33 \text{ m/s} \) to \( 0 \text{ m/s} \) in \( 1.40 \text{ s} \), resulting in a deceleration of \(-202.38 \text{ m/s}^2 \). This deceleration equates to \(-20.63 \text{ g} \) when expressed in terms of gravitational force.
It’s key to understand that an absolute value of \(20.63 \text{ g} \) means the force is significant, but not more than the \(40 \text{ g}\) claimed in the article, hence the figures stated in the magazine were not consistent.
Correct deceleration understanding helps reinforce the accurate assessment of safety measures and design considerations in high-speed scenarios.