Will the Car Stop in Time to Avoid Hitting a Child in a School Zone?

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Homework Help Overview

The problem involves a scenario where a driver must stop a car to avoid hitting a child in a school zone. The car is initially traveling at a speed of 40 km/h, and the child appears 13 m ahead. The driver has a reaction time of 0.25 seconds and the car decelerates at a rate of 8.0 m/s². Participants are exploring whether the car can stop in time.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • One participant calculates the distance traveled during the driver's reaction time and then the stopping distance after braking. Another participant questions their own understanding of the deceleration phase and acknowledges an error in their initial reasoning.

Discussion Status

The discussion includes various calculations and interpretations of the problem. Some participants are verifying their approaches and results, while others are reflecting on their understanding of the physics involved. There is no explicit consensus on the outcome, but participants are engaging with the problem constructively.

Contextual Notes

Participants are working within the constraints of the problem as stated, including the specified speed limit, distance to the child, and deceleration rate. There is an acknowledgment of assumptions regarding the phases of motion, particularly during the driver's reaction time.

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Homework Statement



The speed limit in a school zone is 40 km/h (about 25 mi/h). A driver sees a child run onto the road 13 m ahead of his car. He applies the breaks, and the car decelerates at a uniform rate of 8.0 m/s^2. If the driver's reaction time is 0.25 s, will the car stop before hitting the child?



Homework Equations



d = volt + 1/2at^2
V^2 = Vo^2 + 2ad



The Attempt at a Solution



40 km/h x [1000 m/1 km] x [1 h/3600 s] = 11 m/s

Vo = 11 m/s
X = ?
Xo = 0
a = -8.0 m/s^2
t = 0.25 s

X - Xo = volt + 1/2at^2
X = (11 m/s)(0.25 s) + 1/2(-8.0 m/s^2)(0.25 s)^2
= 2.5 m

Vo = 11 m/s
V = 0
X = ?
Xo = 2.5 m
a = -8.0 m/s^2

V^2 = Vo^2 + 2a(X - Xo)
X = Xo + [V^2 - Vo^2/2a]
= 2.5 + [0 - (11 m/s)^2/2(-8.0 m/s^2)
= 10 m

Therefore, 2.5 m + 10 m = 12.5 m, and the car will stop in time.

Seems right, can anybody please verify this outcome?
 
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Dunno if I'm right, but this is how I see this problem.

First, I calculate the distance car has traveled in that 0.25 seconds

d=v*t=\frac{100m}{9s}*\frac{1s}{4}=\frac{25m}{9}

After that, the car slows down

v^{2}=v_{0}^{2}+2ad
0=\frac{100}{9}^{2}-2*8d
d=\frac{100*100}{81*2*8}

Now to sum up those 2 distances

d=\frac{25}{9}+\frac{10000}{1296}=\frac{13600}{1296}\approx10.5m

But once again, I don't know if I'm rigth (Although I agree the boy will live :biggrin:)
 
Oh, I see where I went wrong. No deceleration during the first part, and a constant velocity. Thank-you for your help!
 
You're welcome :D
 

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