Can the car stop in time at the stop sign?

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SUMMARY

A car traveling at 90 km/h approaches a stop sign 40 meters away and decelerates at -8 m/s². The calculated stopping distance is 39.1 meters, which is less than the 40 meters available, confirming that the car can stop in time. The position equation x(t) = x₀ + v₀t + ½at² was utilized, and the velocity was derived as v(t) = dx/dt = v₀ + at. The discussion emphasizes the importance of differentiation in solving motion problems.

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  • Understanding of kinematic equations, specifically x(t) = x₀ + v₀t + ½at²
  • Knowledge of differentiation and its application in physics
  • Familiarity with basic physics concepts such as velocity and acceleration
  • Ability to perform calculations involving units of speed (km/h to m/s)
NEXT STEPS
  • Study the application of differentiation in kinematics
  • Learn how to convert speed units from km/h to m/s
  • Explore advanced kinematic equations for varying acceleration
  • Investigate real-world applications of motion equations in automotive safety
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Students studying physics, particularly those focusing on kinematics, as well as educators looking for examples of differentiation in motion problems.

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Homework Statement


A car, traveling at a speed of 90km/h, approaches a stop sign 40m ahead. If the car immediately begins to decelerate at a rate of -8m/s^2, will it be able to stop in time? Justify your response


Homework Equations


d = 1/2at^2 + vt

The Attempt at a Solution



I found time by doing t = v/a and found it to be 25/8s
Then I subbed it in d = 1/2at^2 + vt and got an answer of 39.1m, which is smaller than 40m. So it will be able to stop in time.
But this assignment is all about differentiation, so I should be finding a derivative somewhere... but I don't know where to find a derivative to solve this problem!
Please help~~
 
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Hello,

Well, you've kind of already used differentiation, but you may not know it. You have an equation for position:

x(t) = x_0 + v_0t + \frac{1}{2}at^2

The velocity is found by taking the derivative of this equation:

v(t) = \frac{dx}{dt} = v_0 + at

So when you calculated your time (where did you get t = v/a?) you were using this equation.

Hope this helps.
 

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