Crash with the police car /suvat problem

  • Thread starter Thread starter IB1
  • Start date Start date
  • Tags Tags
    Car Crash
AI Thread Summary
The discussion revolves around a physics problem involving a car trailing an unmarked police car while distracted by a cell phone. The initial conditions state both vehicles are traveling at 110 km/h, with the police car braking at 5.00 m/s² after 2.0 seconds of distraction. Participants analyze the distances traveled during this time to determine the separation between the two cars when attention returns. The calculations involve equalizing the distances traveled by both cars and considering the reaction time before braking. The conversation emphasizes the importance of accurately applying kinematic equations to solve for the final speed at impact.
IB1
Messages
24
Reaction score
0

Homework Statement



You are arguing over a cell phone while trailing an unmarked police car by 25.0 m; both your car and the police car are traveling at 110 km/h. Your argument diverts your attention from the police car for 2.0 s (long enough for you to look at the phone and yell,“I won't do that!”). At the beginning of that 2.0 s, the police officer begins braking suddenly at 5.00 m/s2. (a) What is the separation between the two cars when your attention finally returns? Suppose that you take another 0.400 s to realize your danger and begin braking. (b) If you too brake at 5.00 m/s2, what is your speed (in km/h) when you hit the police car?

Homework Equations



v^2 =u^2 +2as, s=ut +\frac{1}{2} at^2

The Attempt at a Solution



I solved the first part. I get the wrong results for part b) because I think I'm making calculation mistakes. The way how I tried to solve it is by making the distances traveled equal.

From s=ut +\frac{1}{2} at^2 where a=\frac{-5m}{s^2}$ we have that since I start breaking 2.4s later than the police car, I must travel a distance s=u(t-2.4) +\frac{1}{2}at^2 whilst the police car travels a distance s'=ut +\frac{1}{2}at^2. Since we crash, we must have s=s'... this way finding t and s, and from there using v^2 =u^2 +2as we find vP.S. The LaTeX looks weird.
 
Physics news on Phys.org
IB1 said:

Homework Statement



You are arguing over a cell phone while trailing an unmarked police car by 25.0 m; both your car and the police car are traveling at 110 km/h. Your argument diverts your attention from the police car for 2.0 s (long enough for you to look at the phone and yell,“I won't do that!”). At the beginning of that 2.0 s, the police officer begins braking suddenly at 5.00 m/s2. (a) What is the separation between the two cars when your attention finally returns? Suppose that you take another 0.400 s to realize your danger and begin braking. (b) If you too brake at 5.00 m/s2, what is your speed (in km/h) when you hit the police car?

Homework Equations



v^2 =u^2 +2as, s=ut +\frac{1}{2} at^2

The Attempt at a Solution



I solved the first part. I get the wrong results for part b) because I think I'm making calculation mistakes. The way how I tried to solve it is by making the distances traveled equal.

From s=ut +\frac{1}{2} at^2 where a=\frac{-5m}{s^2}$ we have that since I start breaking 2.4s later than the police car, I must travel a distance s=u(t-2.4) +\frac{1}{2}at^2 whilst the police car travels a distance s'=ut +\frac{1}{2}at^2. Since we crash, we must have s=s'... this way finding t and s, and from there using v^2 =u^2 +2as we find v


P.S. The LaTeX looks weird.

Conceptualise!

For the first 2.4 seconds, you continued at constant speed while the Police car braked [note: it didn't break - it only did that after you hit it].

Using your formulae [or a more suitable one] you could calculate the distance you have both travelled, and thus how far you are then behind the PC.
You can also calculate how much faster than the police car you are traveling at that time.
From that time on, both you and the PC have the same acceleration, so you will continue to be traveling that same amount faster than the police car, until it, then eventually you, stops - if you haven't already collided.

Let's invent some number - apologies if any of them are actually the correct ones.

Lets assume the initial speed is 30 m/s [it is actually a bit faster]
In 2.4 seconds you will travel 72m
In 2.4 seconds the police car will have slowed to 18 m/s, so average speed is 24 m/s so it has progressed 57.6 m
You have caught up 14.4 m so are now only 10.6m behind.
You are traveling 12 m/s faster than the police car at all times while braking [you each have the same acceleration] so it takes a fraction of a second to catch the Police Car, your speed will have reduced by only perhaps 4 m/s, so you will be traveling quite fast.
Note that your closing speed will only be 12 m/s, so the impact damage won't be too great, but the "spinning off into the scenery" that both you and the Police Car are about to begin could be exciting

NOTE: You might want to rework the problem without the phone, and just include the 0.400 seconds reaction time to see if you stood a chance of stopping before hitting the Police Car anyway! 25m is about 3 car lengths, and on a 110km/h freeway is probably too close to be following anyway!
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top