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

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SUMMARY

The discussion centers on a physics problem involving a car's ability to stop before hitting a child in a school zone. The car, traveling at 40 km/h (11 m/s), has a reaction time of 0.25 seconds and decelerates at a rate of 8.0 m/s². Calculations show that the car travels 2.5 meters during the reaction time and an additional 10 meters while decelerating, totaling 12.5 meters, which is sufficient to stop before reaching the child who is 13 meters away. Therefore, the conclusion is that the car will stop in time.

PREREQUISITES
  • Understanding of kinematic equations, specifically d = vt + 1/2at² and V² = Vo² + 2ad.
  • Knowledge of unit conversions, particularly converting km/h to m/s.
  • Familiarity with concepts of acceleration and deceleration in physics.
  • Basic grasp of reaction time and its impact on motion calculations.
NEXT STEPS
  • Study the application of kinematic equations in real-world scenarios.
  • Learn more about the effects of reaction time on stopping distances in driving safety.
  • Explore advanced topics in physics such as dynamics and motion under constant acceleration.
  • Investigate the implications of speed limits in urban planning and child safety measures.
USEFUL FOR

This discussion is beneficial for physics students, driving instructors, traffic safety analysts, and anyone interested in understanding the dynamics of vehicle stopping distances in emergency situations.

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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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