Physics/Calculus Help: Stopping Distance Calculation for a Decelerating Vehicle

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In summary, a motorist traveling at a speed of 90 feet per second decelerates at a constant rate of 30 feet per second and stops after traveling a distance of 135 feet. This can be calculated using the kinematic motion equations and the formula Vf^2=Vi^2+2ad, where Vf is 0, Vi is 90 feet per second, and a is -30 feet per second.
  • #1
calculusguru
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1. A motorist who was traveling at a speed of 90 feet per second, sees a traffic pile-up ahead and begins to decelerate at a constant rate of 30 feet per second. How war will the vehicle travel before it comes to a complete stop?



2. v=ds/dt
a=dv/dt



3.da/dt=-30
dv/dt=-30t+C
C=90
dv/dt=-30t+90
t=3 when velocity is 0 however I am stuck at how to get the distance traveled
 
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  • #2
calculusguru said:
1. A motorist who was traveling at a speed of 90 feet per second, sees a traffic pile-up ahead and begins to decelerate at a constant rate of 30 feet per second. How war will the vehicle travel before it comes to a complete stop?



2. v=ds/dt
a=dv/dt



3.da/dt=-30
you mean a = dv/dt = -30 (given)
dv/dt=-30t+C
C=90
watch you calculus...dv/dt = -30
dv/dt=-30t+90
when you integrate, you get v = -30t + C, C =90
t=3 when velocity is 0 however I am stuck at how to get the distance traveled
Now you can integrate your first equation, ds = vdt, where v = -30t +90, and solve for s. Note that once you use calculus to derive the kinematic motion equations that relate v, t, and s, and a, for constant acceleration, then you should memorize them, because you'll use them over and over again. In general, specific equations used to solve specific problems should not be memorized, but there are exceptions when using generally applied equations.
 
  • #3
you can always use Vf^2=Vi^2+2ad
Vf would be 0 because it means that the motorcycle has stopped.
Vi would be the initial velocity.
a would be the acceleration.
Solve for d.
 

FAQ: Physics/Calculus Help: Stopping Distance Calculation for a Decelerating Vehicle

What is the difference between physics and calculus?

Physics is the study of matter, energy, and their interactions, while calculus is a branch of mathematics that deals with rates of change and accumulation. In simple terms, physics is the application of calculus to real-world problems.

How can calculus help in understanding physics?

Calculus provides the necessary tools and techniques to analyze and solve complex problems in physics. It helps in determining rates of change, finding maximum and minimum values, and understanding the behavior of physical systems.

What are the key concepts in calculus that are important for physics?

The key concepts in calculus that are important for physics include derivatives, integrals, limits, and differential equations. These concepts help in understanding the fundamental laws and principles of physics, such as Newton's laws of motion and the laws of thermodynamics.

How can I improve my understanding of calculus and physics?

To improve your understanding of calculus and physics, it is important to practice regularly and work through problems. You can also seek help from a tutor or join study groups to discuss and clarify concepts. Additionally, using visual aids and real-world examples can also aid in understanding these subjects.

Are there any real-life applications of calculus in physics?

Yes, there are many real-life applications of calculus in physics. For example, calculus is used in understanding the motion of objects, calculating the speed and acceleration of vehicles, and predicting the behavior of complex systems like weather patterns and fluid flow. It is also essential in the fields of engineering, astronomy, and economics.

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