Calculating Time for Cockroaches to Meet with Different Accelerations

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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
 
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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*
 
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