Two cars travel in the same direction along a straightroad. When a stopwatch reads t=0.00h, car A is atdA=48.0km moving at a constant 36.0km/h. Later,when the watch reads t=0.50h, car B is at dB=0.00kmmoving at 48.0 km/h. Answer the following questions: first,graphically by creating a position-time graph; second algebraicallyby writing down equations for the postion dA anddB as a function of the stopwatch time, t.a) what will the watch read when car B passes car A?b) at what position will the passing occur?c) when the cars pass, how long will it have been since car Awas at the reference point, d= 0 km?

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Alyvia Marks · Accepted Answer

Note that distance is a function of—or is dependent on time.You can plot this on graph paper using time as the x-axis anddistance as the y axis. Algebraically, you can use the point-slopeequation of a line:y - y1 = m(x - x1)where m is the slope,                     y        2            −              y        1                            x        2            −              x        1            Since car A&#039;s initial position (at time zero) is 48km, thefirst point is P1(0, 48). An hour later, the car has traveled 36km,so the point will be P2(1, 48 + 36) = (1, 84). Thus the equation ofthe line becomes:  y  −  48  =            84      −      48              1      −      0        (  x  −  0  )  =  36  x  −  48(or f(t) = 36t + 48)This makes sense because the y-intercept is 48 and for everyhour that passes (or increment in x), the car travels 36km.For car B, the the first point is P3(0.5, 0). After an hour,the car has traveled 48km, so its second point is P4(1.5, 48).Using the same equation you&#039;ll get:  y  −  0  =            48      −      0              1.5      −      5        (  x  −  5  )  =  48  x  −  24(or f(t) = 48t - 24)a) Car B will pass car A when both have traveled the samedistance. This can be found by setting each equation equal to eachother and solving for t:36t + 48 = 48t - 24 =72 = 12tt = 6 sb) Both cars will be in the same position at this time, so youcan plug 6 in for &quot;t&quot; in either and solve:36(6) + 48 = 264 km

embelurildmixjm · Accepted Answer

c) Since the stopwatch read 0.00h when car A was already at 48km, you can set the equation equal to zero (that is, the car has nodistance), but you&#039;ll get a negative time, so you&#039;ll need theabsolute value:36t + 48 = 036t = -48t = -4/3t = 4/3So since &quot;0 km&quot;, car A has been traveling for 6 + 4/3 hours,or 7 1/3 hours.

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