Calculating Stopping Time of Car at 77 km/h

  • Thread starter Thread starter swimstar
  • Start date Start date
  • Tags Tags
    Speed
AI Thread Summary
To calculate the stopping time of a car traveling at 77 km/h (converted to 21.39 m/s) when stopping over a distance of 103 m, the equations for uniformly accelerated motion should be applied. The initial velocity is known, and the average velocity must be used instead of just the initial velocity for accurate calculations. The key focus is on the distance of 103 m, as the problem specifically asks for the stopping time within that distance. The hint about the cat being negligible serves to clarify that the distance to consider is solely the stopping distance. Ultimately, the solution involves determining the time taken to stop using the correct distance and average velocity.
swimstar
Messages
6
Reaction score
0
A driver in a car traveling at a speed of 77 km/h sees a cat 107 m away on the road. How long will it take for the car to accelerate uniformly to stop in exactly 103m? Answer in units of s.


I really have no idea where to begin with this problem.
 
Physics news on Phys.org
Any problem dealing with uniform acceleration starts with the equations for uniformly accelerated motion. Surely these are in your class notes or textbook.

Also, writing out the known quantities (vinitial=___, etc.) would help with this and most other physics problems.
 
I know that I need to convert 77 km/h to 21. 38888889 m/s which is the initial Velocity.

I was thinking of finding T with the equation d=V(initial)*T. So i could therefore put it in the question in: d=V(initial)*T + 1/2at^2 and solve for a.
However, I am unsure if I use 103m or 107 m in the equations.

Am I doing this correct?
 
swimstar said:
However, I am unsure if I use 103m or 107 m in the equations.
Hint: cat is negligible :wink:. This is just a trap. To confuse you what distance to use. And make problem more dramatic.

regards
 
Last edited:
swimstar said:
I know that I need to convert 77 km/h to 21. 38888889 m/s which is the initial Velocity.

I was thinking of finding T with the equation d=V(initial)*T.
That is almost the right equation, except that you need v_average, not v_initial. For v_average you may take the average of the initial and final velocities.
So i could therefore put it in the question in: d=V(initial)*T + 1/2at^2 and solve for a.
You could do that, but look again at what the question is asking you to find. (It is not asking for the acceleration.)
However, I am unsure if I use 103m or 107 m in the equations.
The question asks how long it takes for the car to stop in 103 m.
 
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 'A cylinder connected to a hanging 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...

Similar threads

Back
Top