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Physics 8 Online
OpenStudy (khally92):

Jack is driving with a pail of water along a straight pathway at a steady 25ms when he passes jill who is parked in her minivan waiting for him. when jack is beside jill, she begins accelerating at the rate of 0.004ms^2 in the same direction that jack is driving. how long does it take jill to catch up to jack............. I have solve something similar but here, the acceleration of jack is not given? am i not meant to equate their distance S=UT+1/2at^2.. thanks

OpenStudy (khally92):

help then

OpenStudy (mrdoe):

jack has no acceleration, he is moving with a constant velocity so all you have to do is use your kinematic equations. Since jack has no acceleration, using the position formula we end up with x=0+v(0)t+0 where v(0) would be his constant 25 m/s Then take jill into account, since we count them both as starting at position "0" we get: x=0+v(0)t+1/2at^2 And in this case she has no initial velocity so you can cancel that out, now equate the two expressions to find when they meet (when the two equations are equal), you will have to use the quadratic formula to get the final answer since this will give you a binomial.

OpenStudy (khally92):

I do not undetstant the papameters u used wat is X and 0 in the equation, could u please write it out agin using the S for distance t time, a aceeleleration u velocity and v fot final velocity thanks

OpenStudy (mrdoe):

sure, so using those variables, jack is represented by s=0+ut+0 and jill would be s=0+0+1/2at^2 This is because jack has no acceleration and jill has no initial velocity, and we consider them both to be starting at position "0". From here set them equal to each other and simplify. You should end up with a single binomial expression so all you have to do is use the quadratic formula to solve for t

OpenStudy (khally92):

where did the zeros comes from.........

OpenStudy (khally92):

i.e 25t=0.002t^2

OpenStudy (mrdoe):

for jack's part, the zeros are his initial position and acceleration*time^2, both of which are zero. for jill's part, the zeros are her initial position and velocity*time, again both zero.

OpenStudy (mrdoe):

right, just make it one expression to use the quadratic formula

OpenStudy (khally92):

i.e =0.002t^2-25t=0

OpenStudy (mrdoe):

looks good but i haven't done out the problem myself

OpenStudy (khally92):

woahi got t=0 and t= 12500 is tht possible could u please try it thanks

OpenStudy (mrdoe):

I am getting the same result but if you think about it, with such a small acceleration it should be taking a long time for jill to catch up

OpenStudy (khally92):

well that about 208minutes

OpenStudy (khally92):

thanks alot appreciate

OpenStudy (mrdoe):

ne nada, gl

OpenStudy (khally92):

sorry

OpenStudy (khally92):

thank mr Doe

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