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

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SUMMARY

The discussion focuses on calculating the stopping distance of a vehicle decelerating from 90 feet per second at a constant rate of 30 feet per second. The key equations involved include the relationship between velocity, acceleration, and distance, specifically using the formulas v = ds/dt and a = dv/dt. The integration of these equations leads to the conclusion that the stopping distance can be derived using the kinematic equation Vf² = Vi² + 2ad, where Vf is the final velocity (0), Vi is the initial velocity (90 feet per second), and a is the acceleration (-30 feet per second squared).

PREREQUISITES
  • Understanding of basic calculus concepts, including integration and differentiation.
  • Familiarity with kinematic equations in physics.
  • Knowledge of velocity and acceleration definitions.
  • Ability to solve differential equations related to motion.
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  • Learn how to derive kinematic equations for constant acceleration.
  • Study the application of calculus in physics, particularly in motion problems.
  • Explore advanced integration techniques for solving differential equations.
  • Practice solving real-world problems involving deceleration and stopping distances.
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This discussion is beneficial for physics students, calculus learners, and anyone interested in understanding the mathematical principles behind motion and stopping distances in vehicles.

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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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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.
 
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.
 

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