Homework Help: Kinematic Question -- Car braking to avoid hitting a pedestrian

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1. Mar 6, 2016

Erin Pashanov

1. The problem statement, all variables and given/known data
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=?
2. Relevant equations
d=Vit+½at²
3. 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

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?

2. Mar 6, 2016

drvrm

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 ....

3. Mar 6, 2016

PhanthomJay

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.

4. Mar 6, 2016

Erin Pashanov

Kinda confused.. what should I solve for then? Vf?

5. Mar 6, 2016

PhanthomJay

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.

6. Mar 6, 2016

Erin Pashanov

Got 135.2, now I sub this as my distance?

7. Mar 6, 2016

haruspex

135.2m is the distance for which your motion will be ... what? What will your motion be prior to that?