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?

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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?

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Answer 1 · 2018-03-06T18:10:11

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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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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?

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