How Long Until Two Drag Racing Cars Meet?

  • Thread starter Thread starter i_heart_you
  • Start date Start date
  • Tags Tags
    Cars Drag Racing
AI Thread Summary
Two drag racing cars are analyzed, with the first car traveling at a constant velocity of 40 m/s and starting 10 m ahead of the second car, which accelerates at 3.6 m/s². The problem involves determining the time and displacement until both cars meet. Initial attempts to solve the problem resulted in incorrect equations and negative time values. The correct approach involves using kinematics equations for each car's displacement and setting them equal to find the meeting point. The discussion concludes with the user successfully resolving the problem by correctly calculating the displacements.
i_heart_you
Messages
5
Reaction score
0
2 cars drag racing question...

Homework Statement


The first car has a constant velocity of 40 m/s and moves 10 m before the 2nd car starts moving. The second car speeds up with an acceleration of 3.6 m/s^2. What is the time and the displacesment before the 2 cars actually meet?


Homework Equations


V=d/t
d=V1t x (1/2) a t^2.. i think?

The Attempt at a Solution


I figured out the time it took for the first car to travel 40 m... which would be equal to initial t for the second car. Since I want to find when the two cars have an equal displacement i set their displacements equal and it didn't really work out cause i ended up with a negative time for some reason?
 
Physics news on Phys.org


Hi i_heart_you, welcome to PF.

Start by writing two kinematics equations, one for each car, giving the displacement as a function of time. The first one you posted is inapplicable here because it works for constant velocity only and the second one is incorrect. Don't guess, look 'em up.
 


thanks for your help... i finally figured it out
i just had to split up the displacements into initial and final and make the final displacements equal to each other...
 
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...
Back
Top