Physics, 11.07.2019 15:00 sophiaa23

Anna is driving from champaign to indianapolis on i-74. she passes the prospect ave. exit at noon and maintains a constant speed of 75 mph for the entire trip. chuck is driving in the opposite direction. he passes the brownsburg, in exit at 12: 30pm and maintains a constant speed of 65 mph all the way to champaign. assume that the brownsburg and prospect exits are 105 miles apart, and that the road is straight. how far from the prospect ave. exit do anna and chuck pass each other?

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Answer from: Ancelingabaut

The distance from the prospect ave. where Anna and Chucks pass each other is 73.5 miles

The given parameters;

speed of Anna = Va = 75 mphspeed of Chucks, Vc = 65 mphtime traveled by Anna before Chucks start = 30 mins = 0.5 hourthe distance between Anna and chucks = 105 miles

The distance traveled by Anna before Chucks start is calculated as;

$d_1 = 0.5 \times 75= 37.5 \ miles$

The new distance between Anna and Chucks when Chucks started moving;

$d = 105 - d_1\\\\d = 105 - 37.5 = 67. 5 \ miles$

The time when the two meet is calculated by applying relative velocity formula;

$(V_a - (V_c)) \times t = new \ distance \ between \ them\\\\(V_a + V_c) t = 67.5\\\\(75 + 65) t = 67.5\\\\140 t = 67.5\\\\t = \frac{67.5}{140} \\\\t = 0.48 \ hour$

The distance form the prospect ave. exit when Anna and Chucks passes each other;

This distance will be equal to the initial distance traveled by Anna and distance to where she passes Chucks

$d_2 = d_1 + 0.48(75)\\\\d_2 = 37.5 + 36\\\\d_2 = 73.5 \ miles$

Thus, the distance from the prospect ave. where Anna and Chucks pass each other is 73.5 miles

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Answer from: fluffy374747

Let they both meet after time "t" after Chuck leaves the city

So Monica will move for time "t + 0.5" h as she leave before half an hour

so in this time the sum of distance covered by both of them must be equal to the total distance between them initially

so we have

v_1*(t+0.5) + v_2*t = d

75*t + 37.5 + 65*t = 105

140*t = 67.5

t = 0.48 h

so the total time after which they both will meet is 0.48 hours or 28.9 minutes

Now the distance of the point of intersection from Prospects is given as

d = v_1 * (t + 0.5)

d = 75 * (0.48 + 0.5) = 73.5 miles

so the distance from Prospects is 73.5 miles

Answer from: Quest

t. w e n t y c h a r a c t e r s

Answer from: Quest

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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