Calculating Time for Cockroaches to Meet with Different Accelerations

  • Thread starter Thread starter chonny
  • Start date Start date
  • Tags Tags
    Kinematics
AI Thread Summary
Two cockroaches, initially 60 cm apart, accelerate toward each other with rates of 0.20 m/s² and 0.12 m/s². Using the formula for distance, the total distance covered by both cockroaches can be expressed as the sum of their individual distances. Setting up the equation 0.1t² + 0.06t² = 60 leads to the solution for time, which is approximately 19.36 seconds. The discussion emphasizes the importance of problem-solving in physics and encourages a supportive environment. The calculation concludes with a reminder of the value of life amidst academic stress.
chonny
Please help me...I've been struggling for a long time over this following problem...

Some cockroaches can run as fast as 1.5 m/s. Suppose that two cockroaches are separated by a distance of 60 cm and that they begin to run toward each other at the same moment. Both insects have constant acceleration until they meet. The first cockroach has an acceleration of 0.20 m/s^2 in one direction and the second one has an acceleration of 0.12 m/s^2 in the opposite direction. How much time passes before the two insects bump into each other?


Please give me the answer and the steps leading to the answer as soon as possible...I am most grateful :wink:

If not...I will kill myself
 
Physics news on Phys.org
a_1=0.2
a_2=0.12
v_01=v_02=0
x=x_0+v_0*t+at^2/2
x_1=0+0+a_1*t^2/2=0.2*t^2/2
x_2=0+0+a_2*t^2/2=0.12*t^2/2
x_1+x_2=60
0.2*t^2/2+0.12*t^2/2=60
t=19.36 sec


p.s
dont kill yourself :smile:
 
thank you ...*puts down shotgun*
 
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