Respuesta :
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
[tex]v_1*(t+0.5) + v_2*t = d[/tex]
[tex]75*t + 37.5 + 65*t = 105[/tex]
[tex]140*t = 67.5[/tex]
[tex]t = 0.48 h[/tex]
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
[tex]d = v_1 * (t + 0.5) [/tex]
[tex]d = 75 * (0.48 + 0.5) = 73.5 miles[/tex]
so the distance from Prospects is 73.5 miles
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 mph
- speed of Chucks, Vc = 65 mph
- time traveled by Anna before Chucks start = 30 mins = 0.5 hour
- the distance between Anna and chucks = 105 miles
The distance traveled by Anna before Chucks start is calculated as;
[tex]d_1 = 0.5 \times 75= 37.5 \ miles[/tex]
The new distance between Anna and Chucks when Chucks started moving;
[tex]d = 105 - d_1\\\\d = 105 - 37.5 = 67. 5 \ miles[/tex]
The time when the two meet is calculated by applying relative velocity formula;
[tex](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[/tex]
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
[tex]d_2 = d_1 + 0.48(75)\\\\d_2 = 37.5 + 36\\\\d_2 = 73.5 \ miles[/tex]
Thus, the distance from the prospect ave. where Anna and Chucks pass each other is 73.5 miles
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