What is the Minimum Distance Between a BMW and a Camper to Avoid Collision?

  • Thread starter Thread starter jojo711
  • Start date Start date
AI Thread Summary
To avoid a collision between the BMW and the camper, the driver must account for both his reaction time and braking distance. During the 0.26 seconds of reaction time, the BMW continues to travel at 39.2 m/s, covering a distance of approximately 10.2 meters. After reacting, the driver decelerates at 3.5 m/s², which requires calculating the distance needed to stop before reaching the camper. The camper's speed of 15.7 m/s must also be factored in to determine the minimum safe distance. Properly analyzing both phases of the situation is essential to find the correct minimum distance to avoid a collision.
jojo711
Messages
27
Reaction score
0

Homework Statement


Suppose that a man driving a BMW zooms out of a darkened tunnel at 39.2 m/s. He is momentarily blinded by the sunshine. When he recovers, he sees that he is fast overtaking a camper ahead in his lane moving at the slower speed of 15.7 m/s. He hits the brakes as fast as he can (his reaction time is 0.26 s). If he can decelerate at 3.5 m/s2, what is the minimum distance between the driver and the camper when he first sees it so that they do not collide?

Homework Equations


V=Vo+at
X=Xo+Vot+1/2at^2
V^2=Vo^2+2a(X-Xo)

The Attempt at a Solution


I took the second equation and plugged in the numbers [V=39.2(0.26)+1/2(3.5)(0.26^2)] and ended with the answer being 10.3103. When I put my answer in, it was wrong. Please help me!
 
Physics news on Phys.org
jojo711 said:

Homework Statement


Suppose that a man driving a BMW zooms out of a darkened tunnel at 39.2 m/s. He is momentarily blinded by the sunshine. When he recovers, he sees that he is fast overtaking a camper ahead in his lane moving at the slower speed of 15.7 m/s. He hits the brakes as fast as he can (his reaction time is 0.26 s). If he can decelerate at 3.5 m/s2, what is the minimum distance between the driver and the camper when he first sees it so that they do not collide?

Homework Equations


V=Vo+at
X=Xo+Vot+1/2at^2
V^2=Vo^2+2a(X-Xo)

The Attempt at a Solution


I took the second equation and plugged in the numbers [V=39.2(0.26)+1/2(3.5)(0.26^2)] and ended with the answer being 10.3103. When I put my answer in, it was wrong. Please help me!

Your chosen equation does not address the problem as stated. The 0.26s time span is the driver's reaction time -- the time between when he recognizes the danger and when he actually does something about it. No braking is happening during that time! No velocity change is taking place during that time!

You'll have to consider the problem in two parts: the 'reaction time phase' and the 'braking phase'.
 
Hint: The speed of the BMW must just equal the speed of the camper just the split second it catches it. That defines a time.

As an aside, the brakes on the BMW are not working properly. The average vehicle has an acceleration rate under maximum braking of -1g which is -9.8 m/s^2.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top