A car travelling at \(36 \mathrm{~km} / \mathrm{h}\) crashes into a mountainside. The crunchzone of the car deforms in the collision, so that the car effectively stops over a distance of \(1 \mathrm{~m}\). (a) Let us assume that the acceleration is constant during the collision, what is the acceleration of the car during the collision? (b) Compare with the acceleration of gravity, which is \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\).

Short Answer

Expert verified
The car's acceleration during the collision is \( -50 \text{ m/s}^2 \), approximately 5.1 times greater than gravity.

Step by step solution

01

Convert Velocity to m/s

First, convert the given velocity from km/h to m/s. Use the conversion factor: \[ \text{Velocity (m/s)} = \text{Velocity (km/h)} \times \frac{1000}{3600} \] \[ 36 \mathrm{~km/h} = 36 \times \frac{1000}{3600} \text{ m/s} = 10 \text{ m/s} \]
02

Use the Kinematic Equation

Use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance: \[ v^2 = u^2 + 2as \] Here, final velocity (\[v\] = 0 m/s), initial velocity (\[ u = 10 \] m/s), and stopping distance (\[s = 1 \] m). We need to find acceleration ( \[a\]). Rearrange the equation to solve for acceleration: \[ 0 = 10^2 + 2a \times 1 \]
03

Solve for Acceleration

Simplify and solve for acceleration: \[ 0 = 100 + 2a \] \[ -100 = 2a \] \[ a = -50 \text{ m/s}^2\] The negative sign indicates that the acceleration is in the opposite direction of the car’s motion.
04

Compare with Gravity

Compare the calculated acceleration with the acceleration due to gravity \(g = 9.8 \text{ m/s}^2 \): \[ \left| \frac{a}{g} \right| = \left| \frac{-50}{9.8} \right| \approx 5.1 \] The magnitude of the car’s acceleration is approximately 5.1 times the acceleration due to gravity.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

kinematic equations
Kinematic equations are fundamental in physics, especially when dealing with motion. They relate various parameters like velocity, acceleration, time, and displacement without involving mass or force. One of the most used equations is: \(v^2 = u^2 + 2as\). In this equation:
  • \textbf{v}: final velocity
  • \textbf{u}: initial velocity
  • \textbf{a}: acceleration
  • \textbf{s}: displacement
This equation helped us find the acceleration of the car during the collision. Having the final velocity (\text{v}) as 0, initial velocity (\text{u}) as 10 m/s, and stopping distance (\text{s}) as 1 m, we rearranged the equation to solve for the unknown acceleration (\text{a}). This approach is common in problems where constant acceleration is assumed.
acceleration
Acceleration is the rate at which an object's velocity changes over time. It involves both the magnitude and direction, making it a vector quantity. In the problem, the acceleration was calculated using the kinematic equation. Here's a quick recap of the calculation:
  • Start with the kinematic equation: \(v^2 = u^2 + 2as\).
  • Since the car stops, \text{v} is 0: \(0 = 10^2 + 2a \times 1\).
  • Rearrange to isolate \text{a}: \( -100 = 2a \rightarrow a = -50 \text{ m/s}^2 \).
The negative sign is important as it indicates deceleration (negative acceleration), meaning the car is slowing down.
velocity conversion
In physics, using consistent units is crucial. The exercise gave the car's speed in \text{km/h}, which we had to convert to \text{m/s} to use in the kinematic equations. The conversion factor is: \( \text{Velocity (m/s)} = \text{Velocity (km/h)} \times \frac{1000}{3600} \). For this problem: \( 36 \text{ km/h} = 36 \times \frac{1000}{3600} \text{ m/s} = 10 \text{ m/s} \). This straightforward conversion ensures our calculations are accurate and consistent with the SI unit system, which is typically used in physics.
comparison with gravity
After finding the car's acceleration during the collision, it's helpful to compare it with the acceleration due to gravity, \text{g} (which is \( 9.8 \text{ m/s}^2 \)). Gravity is a constant force acting on objects near Earth's surface. Here, we found:
  • Car's acceleration: \( -50 \text{ m/s}^2 \)
  • Acceleration due to gravity: \( 9.8 \text{ m/s}^2 \)
To compare: \( \frac{|a|}{g} = \frac{|-50|}{9.8} \approx 5.1 \). This means the car experienced a deceleration about 5.1 times stronger than gravity, indicating the severity of the crash.

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Most popular questions from this chapter

An electron is shot through a box containing a constant electric field, getting accelerated in the process. The acceleration inside the box is \(a=2000 \mathrm{~m} / \mathrm{s}^{2}\). The width of the box is \(1 \mathrm{~m}\) and the electron enters the box with a velocity of \(100 \mathrm{~m} / \mathrm{s}\). (a) What is the velocity of the electron when it exits the box?

As an expert archer you are able to fire off an arrow with a maximum velocity of \(50 \mathrm{~m} / \mathrm{s}\) when you pull the string a length of \(70 \mathrm{~cm}\). (a) If you assume that the acceleration of the arrow is constant from you release the arrow until it leaves the bow, what is the acceleration of the arrow?

Your roommate sets off early to school, walking leisurely at \(0.5 \mathrm{~m} / \mathrm{s}\). Thirty minutes after she left, you realize that she forgot her lecture notes. You decide to run after her to give her the notes. You run at a healthy \(3 \mathrm{~m} / \mathrm{s}\). (a) What is her position when you start running? (b) What is your position when \(tt_{1}\). (k) How can you use this result to find where you catch up with your roommate? (1) Where do you catch up with your roommate? (m) What parts of your solution strategy are general, that is, what parts of your strategy do not change if we change how either person moves?

In this project we address the motion of an object sliding on a slippery surface-such as a ski sliding in a snowy track. You will learn how to find the equation of motion for sliding systems both analytically and numerically, and to interpret the results. We start by studying a simplified situation called frictional motion: A block is sliding on a surface. moving with a velocity \(v\) in the positive \(x\)-direction. The forces from the interactions with the surface results in an acceleration: $$ a=\left\\{\begin{array}{cc} -\mu(|v|) g & v>0 \\ 0 & v=0 \\ \mu(|v|) g & v<0 \end{array}\right. $$ where \(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\) is the acceleration of gravity. Let us first assume that \(\mu(v)=\) \(\mu=0.1\) for the surface. That is, we assume that the coefficient of friction does not depend on the velocity of the block. We give the block a push and release it with a velocity of \(5 \mathrm{~m} / \mathrm{s}\). (a) Find the velocity, \(v(t)\), of the block. (b) How long time does it take until the block stops? (c) Write a program where you find \(v(t)\) numerically using Euler's or Euler- Cromer's method. (Hint: You can find a program example in the textbook.) Use the program to plot \(v(t)\) and compare with your analytical solution. Use a timestep of \(\Delta t=0.01 .\) The description of friction provided above is too simplified. The coefficient of friction is generally not independent of velocity. For dry friction, the coefficient of friction can in some cases be approximated by the following formula: $$ \mu(v)=\mu_{d}+\frac{\mu_{s}-\mu_{d}}{1+v / v^{*}} $$ where \(\mu_{d}=0.1\) often is called the dynamic coefficient of friction, \(\mu_{s}=0.2\) is called the static coefficient of friction, and \(v^{*}=0.5 \mathrm{~m} / \mathrm{s}\) is a characteristic velocity for the contact between the block and the surface. (d) Show that the acceleration of the block is: $$ a(v)=-\mu_{d g}-g \frac{\mu_{s}-\mu_{d}}{1+v / v^{*}} $$ for \(v>0\). (e) Use your program to find \(v(t)\) for the more realistic model, with the same starting velocity, and compare with your previous results. Are your results reasonable? Explain. The model we have presented so far is only relevant at small velocities. At higher velocities the snow or ice melts, and the coefficient of friction displays a different dependency on velocity: $$ \mu(v)=\mu_{m}\left(\frac{v}{v_{m}}\right)^{-\frac{1}{2}} \text { when } v>v_{m} $$ where \(v_{m}\) is the velocity where melting becomes important. For lower velocities the model presented above with static and dynamic friction is still valid. (f) Show that $$ \mu_{m}=\mu_{d}+\frac{\mu_{s}-\mu_{d}}{1+v_{m} / v^{*}}, $$ in order for the coefficient of friction to be continuous at \(v=v_{m}\). (g) Modify your program to find the time development of \(v\) for the block when \(v_{m}=1.5 \mathrm{~m} / \mathrm{s}\). Compare with the two other models above: The model without velocity dependence and the model for dry friction. Comment on the results. (h) The process may be clearer if you plot the acceleration for all the three models in the same plot. Modify your program to plot \(a(t)\), plot the results, and comment on the results. What would happen if the initial velocity was much higher or much lower than \(5 \mathrm{~m} / \mathrm{s}\) ?

Is is possible to envision a motion where you for a period have no displacement, but non-zero velocity? (You may use an \(x\) ( \(t\) ) plot for illustration).

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