When a stopwatch reads t = 0.00 h , car A is at d ( a ) =48.0km moving at a consant 36.0 k m/ h . Later, when the watch reads t = 1.50 h , car B is at d ( b ) = 0.00 k m moving at 48.0 k m/ h . Create a position-time graph and algebraic equation?

1 Answer
Jane
Mar 6, 2018

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From the diagram, at #t=0# car #A# has is at #48Km#,it moves with constant velocity = slope of displacement -time curve=#tan u=36#,so its equation of motion is #S=36t+48#,to cross check put #t=0# in the equation,you get, #S=48Km#(displacement when time counting was started i.e #t=0#)

And, at #t=1.5# car #B# starts journe,moving with constant velocity = slope of displacement-time curve=#tan b=48#,so its equation of motion is #S=48(t-1.5)#,to cross check put #t=1.5# you wil get, #S=0#

No,suppose both will meet after time #t# i.e having same displacement,

So,#S_A= S_B#

or, #36t+48=48(t-1.5)#

or, #t=10#

So,when the stop watch reads #10 hrs# both will have same displacement ,i.e #S=408Km#

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