# Solving A-B-C Ship Journey with Motorboat

• jiboom
In summary, the motorboat leaves ship A and travels to ship B in a straight line with a speed of 14 m/s. The time it takes for the motorboat to return back to ship A is 500 seconds.
jiboom
a ship A is traveling due east at a speed of 8m/s and a ship B is traveling due south at 10m/s. At an instant when A is 3km from B in a direction 060 a motor boat leaves A and travels in a straight line to B with speed 14m/s

1) show it reacges B in 500 seconds
2)if upon reaching B the motor boat immediately returns back to A in a straight line,again with speed 14,find the time taken for the return journey.

i need a diagram here to get started. book I am working through is no help :(

im thinking:

distance triangle: if C is point where motorboat meets B then have

AB=3
angle ABC=X say

velocity triangle:

Vb=velocity of B
Vm=velocity of motorboat

then
i have Vb pointing down, -vm at an angle up from bottom of vb and vb-vs makes other side

then angle between vb and vb-vs=180-x

but now not sure how to proceed so thinking my pictures are wrong or I am totally missing something. i seem to need another angle or length?

any takers?

im sitting here with loads of trianlgles on various pieces of paper! I am guessing i get another angle or length from the set up from the question but I am not getting it!

jiboom said:
a ship A is traveling due east at a speed of 8m/s and a ship B is traveling due south at 10m/s. At an instant when A is 3km from B in a direction 060 a motor boat leaves A and travels in a straight line to B with speed 14m/s

1) show it reacges B in 500 seconds
2)if upon reaching B the motor boat immediately returns back to A in a straight line,again with speed 14,find the time taken for the return journey.

i need a diagram here to get started. book I am working through is no help :(

im thinking:

distance triangle: if C is point where motorboat meets B then have

AB=3
angle ABC=X say

velocity triangle:

Vb=velocity of B
Vm=velocity of motorboat

then
i have Vb pointing down, -vm at an angle up from bottom of vb and vb-vs makes other side

then angle between vb and vb-vs=180-x

but now not sure how to proceed so thinking my pictures are wrong or I am totally missing something. i seem to need another angle or length?

it appears what i was attempting was wrong.

i drew a distance diagram showing the point B and M(otorboat) met and then said relative velocity needed to be parallel to the line.

it turns out the relative velocity needs to be parallel to the initial situation, ie at 60 to north, which gives me the extra info i needed to do part (a)

can anyone run this by me though? why is this the way to go

for the second part:

i have tried to find the bearing of A from B and the distance by:

i know boats been moving for 500 seconds so A has moved 4000m right,B 5000m down.

i know when they were 3000 m apart the angle was 60 so i can use this to fill in the top left corner of the right angled triangle.

from this triangle i can find the distance AB and the bearing (roughly 9500,45)

now following the first part, i say relative velocity vector goes at 45 to north, andf in my triangle i have

-va along the base, vm from left hand side of -va and (vm-va) from rhs of -va

and angle between va and (vm-va) =90+45=135

but this does not get me the book answer. I get about 1250 to their 1790.

still no joy matching book answer.

im sure the distance between the 2 is 9500ish number so taking book answer the relative velocity must be 5.2

does this make sense? the relative velocity for the motorboat and B was 6,but B was moving at 10 m.s so should the relative velocity for motorboat and A be greater as A is only moving at 8?

im thinking along lines of:
3 cars on motorway:
car A going 30
car b going 50
car c going 60

relative to A car C is going 30
relative to b car c is going 10
so although B is going quicker its relative velocity is slower

anyone with any thoughts? i have whole exercise of these so can't make any progress til i get this,the first question,sorted. thanks

## 1. How does a motorboat help in solving an A-B-C ship journey?

A motorboat can help in solving an A-B-C ship journey by providing a faster and more efficient mode of transportation compared to a traditional ship. It can also navigate through smaller and more complex waterways, allowing for a more direct route to the destination.

## 2. What factors should be considered when using a motorboat to solve an A-B-C ship journey?

Some factors that should be considered include the distance of the journey, the weather and water conditions, the capacity and capabilities of the motorboat, and the availability of fuel and supplies.

## 3. How does the speed of the motorboat impact the duration of the A-B-C ship journey?

The speed of the motorboat can significantly impact the duration of the A-B-C ship journey. A faster motorboat can cover more distance in a shorter amount of time, reducing the overall duration of the journey. However, it is essential to consider other factors such as weather conditions and safety when determining the speed of the motorboat.

## 4. Can a motorboat be used for all types of A-B-C ship journeys?

While a motorboat can be used for most A-B-C ship journeys, it may not be suitable for all situations. Factors such as the distance, water conditions, and cargo may determine whether a motorboat is the most appropriate mode of transportation for a particular journey.

## 5. How can a motorboat be operated safely during an A-B-C ship journey?

To operate a motorboat safely during an A-B-C ship journey, it is essential to follow all safety precautions and guidelines. This includes wearing appropriate safety gear, understanding and following navigation rules, and regularly maintaining the motorboat to ensure its proper functioning.

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