Hi guys! i'm currently doing homework on Physics and i'm stuck on this question. can anybody help me out ASAP please?



Object A and B start off at the same point at the same time moving along a straight line in the same direction. Object A has a constant velocity of 5.0 m/s while object B has only an acceleration of 1.0m/s-2. When and where will the objects meet again?

thank you

Curlyben

Thank you for taking the time to copy your homework to AMHD.
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ebaines

You need to memorize and learne how to use the equation for position versus time:



For this problem for object A you have = 5 m/s and = 0. For object B you have = 0 and = 1 m/s^2. For both objects you have = 0. When the objects meet both are at the same place - so set the equation for object A equal to the equation for object B- this will allow you to determine the time when they meet. Then shove that value for time back into one of the position equations to get d. Check by putting that same value for time into the other position equation and see if you get the same value for d.

GDss

you see, all i know that is Va=Vb . i don't know which formula to use and this is just a revision not an assignment. but it's okay if you cannot help.

ebaines

No! When the object meet does not equal . Think about this real world example - you are in a car that is stopped on the road when a second car goes past at a steady speed of 30 MPH. You step on the gas, and your car accelerates from zero. At first the other car contines to pull away from you, but once your speed matches his you stop losing ground and can begin to catch up. At that point you are stlll far behind, so even though your speeds match you are not at the same place as him. But your car continues to accelerate, and now that you're going faster than him you start to catch up, and after a bit of time you go rocketing past him at a very high speed. So when you are at the same point as him (as you pass him) your speed is much greater than his, right?

GDss

thank you very much again for your help Ebaines. i really appreciate it. this was just a revision question and not an assignment as i said. thus, thank you for helping me revise this question.
take care.