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For a particular delivery, a courier travels by bicycle at a speed of 
v miles per hour for a distance of 1.3 miles. After making the delivery, the courier travels the same distance back, but travels 2 miles per hour faster than on the way to the delivery. If the courier spent 0.2 hours travelling to and from the delivery, which of the following equations could be used to determine the speed of the bicycle on the way to the delivery?
Choose 1 answer:
(A) 
0.2v^(2)-2.2 v-2.6=0
(B) 
1.3v^(2)-2v-0.2=0
(c) 
0.2(v^(2)-3)+2.6=0
(D) 
3(0.2-v)(2.6-v)=0

For a particular delivery, a courier travels by bicycle at a speed of v v miles per hour for a distance of 11.33 miles. After making the delivery, the courier travels the same distance back, but travels 22 miles per hour faster than on the way to the delivery. If the courier spent 00.22 hours travelling to and from the delivery, which of the following equations could be used to determine the speed of the bicycle on the way to the delivery?\newlineChoose 11 answer:\newline(A) 0.2v22.2v2.6=0 0.2 v^{2}-2.2 v-2.6=0 \newline(B) 1.3v22v0.2=0 1.3 v^{2}-2 v-0.2=0 \newline(C) 0.2(v23)+2.6=0 0.2\left(v^{2}-3\right)+2.6=0 \newline(D) 3(0.2v)(2.6v)=0 3(0.2-v)(2.6-v)=0

Full solution

Q. For a particular delivery, a courier travels by bicycle at a speed of v v miles per hour for a distance of 11.33 miles. After making the delivery, the courier travels the same distance back, but travels 22 miles per hour faster than on the way to the delivery. If the courier spent 00.22 hours travelling to and from the delivery, which of the following equations could be used to determine the speed of the bicycle on the way to the delivery?\newlineChoose 11 answer:\newline(A) 0.2v22.2v2.6=0 0.2 v^{2}-2.2 v-2.6=0 \newline(B) 1.3v22v0.2=0 1.3 v^{2}-2 v-0.2=0 \newline(C) 0.2(v23)+2.6=0 0.2\left(v^{2}-3\right)+2.6=0 \newline(D) 3(0.2v)(2.6v)=0 3(0.2-v)(2.6-v)=0
  1. Define Speed and Distance: Let's denote the speed of the bicycle on the way to the delivery as vv (in miles per hour). The distance for one way is 1.31.3 miles. The time taken to travel to the delivery can be calculated using the formula time=distancespeed\text{time} = \frac{\text{distance}}{\text{speed}}, which in this case is time=1.3v.\text{time} = \frac{1.3}{v}.
  2. Calculate Time for Delivery: On the way back, the courier travels 22 miles per hour faster, so the speed is v+2v + 2. The time taken for the return trip is time=1.3v+2\text{time} = \frac{1.3}{v + 2}.
  3. Formulate Total Time Equation: The total time spent travelling to and from the delivery is given as 0.20.2 hours. Therefore, the sum of the times for both trips should equal 0.20.2 hours. This gives us the equation rac{1.3}{v} + rac{1.3}{v + 2} = 0.2.
  4. Solve for Common Denominator: To solve this equation, we need to find a common denominator and combine the terms. The common denominator for vv and v+2v + 2 is v(v+2)v(v + 2). Multiplying both sides of the equation by v(v+2)v(v + 2) to clear the denominators gives us 1.3(v+2)+1.3v=0.2v(v+2)1.3(v + 2) + 1.3v = 0.2v(v + 2).
  5. Expand and Combine Terms: Expanding both sides of the equation, we get 1.3v+2.6+1.3v=0.2v2+0.4v1.3v + 2.6 + 1.3v = 0.2v^2 + 0.4v. Combining like terms on the left side gives us 2.6v+2.6=0.2v2+0.4v2.6v + 2.6 = 0.2v^2 + 0.4v.
  6. Simplify the Equation: Subtracting 2.6v+2.62.6v + 2.6 from both sides to bring all terms to one side of the equation results in 0=0.2v2+0.4v2.6v2.60 = 0.2v^2 + 0.4v - 2.6v - 2.6. Simplifying the equation by combining like terms gives us 0=0.2v22.2v2.60 = 0.2v^2 - 2.2v - 2.6.
  7. Finalize the Solution: The simplified equation is 0.2v22.2v2.6=00.2v^2 - 2.2v - 2.6 = 0. This matches option (A) from the given choices.

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