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292.982
292.982
incorrect
least one of the answers above is NOT correct.
fishing boat leaves port at 5 miles per hour at a bearing of 
90^(@) for 4 hours, then turns to a bearing of 
260^(@) at 2 miles per hour for 5 hours, and fina hanges to a bearing of 
130^(@) at 9 miles per hour for 5 hours. At this point, the boat heads directly back to port at a speed of 6 miles per hour. Find 
t me it takes the boat to return to port as well as the boat's bearing as it does.
eturn time: 
◻ hours
teturn bearing: 
◻

292292.982982\newline292292.982982\newlineincorrect\newlineleast one of the answers above is NOT correct.\newlinefishing boat leaves port at 55 miles per hour at a bearing of 90 90^{\circ} for 44 hours, then turns to a bearing of 260 260^{\circ} at 22 miles per hour for 55 hours, and fina hanges to a bearing of 130 130^{\circ} at 99 miles per hour for 55 hours. At this point, the boat heads directly back to port at a speed of 66 miles per hour. Find t t me it takes the boat to return to port as well as the boat's bearing as it does.\newlineeturn time: \square hours\newlineteturn bearing: \square

Full solution

Q. 292292.982982\newline292292.982982\newlineincorrect\newlineleast one of the answers above is NOT correct.\newlinefishing boat leaves port at 55 miles per hour at a bearing of 90 90^{\circ} for 44 hours, then turns to a bearing of 260 260^{\circ} at 22 miles per hour for 55 hours, and fina hanges to a bearing of 130 130^{\circ} at 99 miles per hour for 55 hours. At this point, the boat heads directly back to port at a speed of 66 miles per hour. Find t t me it takes the boat to return to port as well as the boat's bearing as it does.\newlineeturn time: \square hours\newlineteturn bearing: \square
  1. Calculate Distance Covered: First leg of the journey: The boat travels at 5mph5 \, \text{mph} for 4hours4 \, \text{hours} on a bearing of 9090 degrees.\newlineCalculate the distance covered: Distance == Speed ×\times Time.\newlineDistance =5miles/hour×4hours=20miles.= 5 \, \text{miles/hour} \times 4 \, \text{hours} = 20 \, \text{miles}.
  2. Calculate Distance Covered: Second leg of the journey: The boat travels at 2mph2\,\text{mph} for 5hours5\,\text{hours} on a bearing of 260260 degrees.\newlineCalculate the distance covered: Distance == Speed ×\times Time.\newlineDistance =2miles/hour×5hours=10miles.= 2\,\text{miles/hour} \times 5\,\text{hours} = 10\,\text{miles}.
  3. Calculate Distance Covered: Third leg of the journey: The boat travels at 9mph9 \, \text{mph} for 55 hours on a bearing of 130130 degrees.\newlineCalculate the distance covered: Distance == Speed ×\times Time.\newlineDistance =9miles/hour×5hours=45miles.= 9 \, \text{miles/hour} \times 5 \, \text{hours} = 45 \, \text{miles}.
  4. Calculate Total Distance: Now, we need to calculate the total distance traveled before heading back to port.\newlineTotal distance = First leg + Second leg + Third leg.\newlineTotal distance = 2020 miles + 1010 miles + 4545 miles = 7575 miles.
  5. Calculate Time to Return: The boat heads directly back to port at a speed of 66 miles per hour.\newlineCalculate the time it takes to return to port: Time == Distance // Speed.\newlineTime == 7575 miles // 66 miles/hour == 12.512.5 hours.

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