What is the maximum speed you could have and still avoid hitting the deer?

  • Thread starter Thread starter xfaceless
  • Start date Start date
  • Tags Tags
    Kinematics Physics
AI Thread Summary
The discussion revolves around calculating the maximum speed a car can travel without hitting a deer that appears 45 meters ahead. The initial speed is 18 m/s, and the driver has a reaction time of 0.5 seconds, leading to a stopping distance of 22.5 meters after braking. For the maximum speed calculation, the total distance covered during the reaction time and braking must be less than 45 meters. A misunderstanding occurred regarding the distance available for braking after accounting for reaction time. Ultimately, the correct approach involves using kinematic equations to determine the maximum initial speed that allows for safe stopping.
xfaceless
Messages
4
Reaction score
0

Homework Statement



You're driving down the highway late one night at 18m/s when a deer steps onto the road 45m in front of you. Your reaction time before stepping on the brakes is 0.50s and the maximum deceleration of your car is 12m/s^2.

a)How much distance is between you and the deer when you come to a stop?
b)What is the maximum speed you could have and still not hit the deer?

Homework Equations



vf^2=vi^2+2ad

The Attempt at a Solution



I figured out part a) by calculating the actual distance the deer is infront of your car (by using reaction time given), and then solved for the distance you travel after you hit the brakes using the kinematics equation.
so, the distance between you & the deer when your car comes to a stop is 22.5m.

For part b), .. I don't really understand part b). I tried solving for vi by putting vf at 0, a=-12, and d=36 but that didn't lead to the right answer. I got vi=29.4m/s which is wrong.

Any help please ? ):
 
Physics news on Phys.org
Assuming the same reaction time, you can use the same equation you used in part a), but this time making the initial speed of the car an unknown. You will want the total distance traveled to be less than the 45m, so as to not hit the deer.

I think in your attempt, you assumed he was going at 18m/s in the distance covered during reaction time.
 
For part b) I did this..
vf^2=vi^2+2ad
0=vi^2+2(-12)(45)
vi=33m/s.
But that's incorrect..
I still dont' get it. :T
 
Now you're giving the car 45m of braking distance - in other words 0 reaction time. If you've got 45m total, and you travel at vi for 0.5s before decelerating, how much distance have you got for braking?

A quick sketch of velocity versus time, and distance versus time can really help to visualise this.
 
Nevermind. I got it.
 
Thanks everyone.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top