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When do they meet?

  • Thread starter Walan
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  • #1
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Homework Statement



Two cars approach each other from opposite directions. One accelerating due east at 100 m/sec^2 and the other accelerating due west at 150 m/sec^2. If they are 500m apart and start from rest, how long does it take for them to collide? Assume constant acceleration.

Homework Equations

. [/B]


The Attempt at a Solution

. [/B]
I can do this with velocities, but am unsure how things change given acceleration. With velocity, one equation was x=100t where x is displacement and t is time, then y=150t where y is displacement and t is also time. The time to impact is the same for each car though the displacement is not. Also, the total displacement of the two cars is 500m so, x + y = 500m
I solved for time and that was that. When I did the same thing with acceleration I got time in sec^2 and knew I had a problem. Would I use the equation x = (Vo)t + 1/2a(t^2) where x is displacement, Vo is initial velocity (which = 0), a is acceleration and t is time? If I used each acceleration in this equation, one displacement x and one displacement y with the total of x + y = 500m would that be right?
 

Answers and Replies

  • #2
ehild
Homework Helper
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Homework Statement



Two cars approach each other from opposite directions. One accelerating due east at 100 m/sec^2 and the other accelerating due west at 150 m/sec^2. If they are 500m apart and start from rest, how long does it take for them to collide? Assume constant acceleration.

Homework Equations

. [/B]


The Attempt at a Solution

. I can do this with velocities, but am unsure how things change given acceleration. With velocity, one equation was x=100t where x is displacement and t is time, then y=150t where y is displacement and t is also time. The time to impact is the same for each car though the displacement is not. Also, the total displacement of the two cars is 500m so, x + y = 500m
I solved for time and that was that. When I did the same thing with acceleration I got time in sec^2 and knew I had a problem. Would I use the equation x = (Vo)t + 1/2a(t^2) where x is displacement, Vo is initial velocity (which = 0), a is acceleration and t is time? If I used each acceleration in this equation, one displacement x and one displacement y with the total of x + y = 500m would that be right?[/B]
Welcome to PF!
As the cars are accelerating, your first approach is wrong. And the given data, 100m/s^2 and 150 m/s2 are accelerations, not velocities.
The second approach is right, go ahead!
 
  • #3
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Thank you for your help with my question!
 

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