Kinematic Question -- Car braking to avoid hitting a pedestrian

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Homework Help Overview

The discussion revolves around a kinematic problem involving a car traveling at a steady speed of 26 m/s that needs to stop to avoid hitting a child 150 m away. The braking acceleration is given as -2.5 m/s², and participants are exploring the implications of reaction time and the necessary calculations to determine stopping distance and time.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss the need to calculate the distance required to stop the car after hitting the brakes, as well as the reaction time before braking. There is mention of using different kinematic equations to relate initial velocity, final velocity, distance, and acceleration.

Discussion Status

Some participants have provided guidance on how to approach the problem, suggesting the calculation of stopping distance and reaction time. There are multiple interpretations of the problem, particularly regarding what variables to solve for, such as final velocity and distance traveled during the reaction time.

Contextual Notes

Participants are grappling with the implications of the physical situation, including the time taken to react before braking and the distance covered during that reaction time. There is uncertainty about the correct approach to take in solving the problem.

Erin Pashanov
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Homework Statement


Driving along a steady speed of 26m/s and suddenly see a child 150m from you. Breaks can produce acceleration of -2.5m/s² but it takes time to get the foot from the gas to the brake pedal. How much time do you have, if to avoid hitting the child?
Known:
d=150m
a=-2.5m/s²
Vi=26m/s
t=?

Homework Equations


d=Vit+½at²

The Attempt at a Solution


150=26(t)+½(-2.5)t²
150=26t+(-1.25)t²
150=26t-1.25t²
1.25t²-26t+150=0
Quadratic formula:
qform01.gif

x=[26±√(-26)²-4(1.25)(150)]/2(1.25)
x=[26±√676-750]/2.5
x=[26±√-74]/2.5
x=[26±8.6i]/2.5
Here I got stuck. How can I solve it if I have an imagery number? I guess my way is wrong?
 
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Erin Pashanov said:
Driving along a steady speed of 26m/s and suddenly see a child 150m from you. Breaks can produce acceleration of -2.5m/s² but it takes time to get the foot from the gas to the brake pedal. How much time do you have, if to avoid hitting the child?
Known:
d=150m
a=-2.5m/s²
Vi=26m/s
t=?

Homework Equations


d=Vit+½at²
i think you are trying to solve without realising the physical situation.
try to ask - what should be car;s deceleration such that it comes to velocity zero after traversing the separation distance ? using other kinematical equation relating initial velocity , final velocity , distance and acceleration you can get info of required acceleration -compare this value with the maximum one can have -then relate the time taken ...
 
The problem is asking about your "reaction" time, that is , the time it takes for you to hit the brakes after first seeing the child. You are traveling at 26 m/s constant speed during this reaction time. You should first calculate how far you will travel to a stop after hitting the brakes , by using another kinematic equation. Then you can calc your max reaction time based on what distance you have left.
 
Kinda confused.. what should I solve for then? Vf?
 
Erin Pashanov said:
Kinda confused.. what should I solve for then? Vf?
No, you should solve for the distance it takes to go from Vo (26 m/s ) to Vf (0 m/s ) when you hit the brakes.
 
PhanthomJay said:
No, you should solve for the distance it takes to go from Vo (26 m/s ) to Vf (0 m/s ) when you hit the brakes.
Got 135.2, now I sub this as my distance?
 
Erin Pashanov said:
Got 135.2, now I sub this as my distance?
135.2m is the distance for which your motion will be ... what? What will your motion be prior to that?
 

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