RELATIVE VELOCITY!!!!

Y_Shun SG

Another QTN.

A man who swims at 1.2m/s in still water wishes to cross a river which is flowing between straight parallel banks at 2m/s. He aims downstream in a direction making an angle of 60° with the bank. Find

the speed at which he travels
the angle which his resultant velocity makes with the bank

Using cosine rule
squareroot
1.2^2 + 2^2 - 2(1.2)(2)cos60°

= 1.743559577

ans is different alot...how cums ar?

EDIT:// PAISEH, READ WRONGLY

lpx88 SG

. diaoz

nightzip SG

so this question is settled already right? Confused

tanjun SG

Originally posted by Y_Shun:
Another QTN.

A man who swims at 1.2m/s in still water wishes to cross a river which is flowing between straight parallel banks at 2m/s. He aims downstream in a direction making an angle of 60° with the bank. Find

the speed at which he travels
the angle which his resultant velocity makes with the bank

Using cosine rule
squareroot
1.2^2 + 2^2 - 2(1.2)(2)cos60°

= 1.743559577

ans is different alot...how cums ar?

[b]EDIT:// PAISEH, READ WRONGLY [/b]

One mistake u have made. From the qn, you have said that the man aims downstream. From the diagram that u have drawn, the man is aiming upstream making an angle of 60 degrees to the bank. Instead of drawing a negative gradient line Vm/w, your Vm/w should be a positive gradient line making an angle of 60 degrees to the bank.

Using cosine rule
square root
1.2^2 + 2^2 - 2(1.2)(2)cos (180 - 60) degrees

= 2.8 m/s

tanjun SG

Not true either. Firstly, your calculator will only give you the smallest value of the angle. Sine is positive between 0 - 180 degrees and hence this occurs. Thus, your diagram is very important in dropping you a hint whether beta is a acute or obtuse angle.

If I use your qn as an example, if beta is an acute angle, the total angle added up will not be 180 degrees which a triangle cannot be formed since x is an acute angle. Thus, by inspection, by logic, beta angle cannot be acute. It must be obtuse. Therefore, your diagram is important as you can see. It is gd that u have drawn the diagram in such a way that x is an acute angle instead of well obtuse angle.

If the qn states that x is an obtuse angle, you will have to draw your diagram differently, drawing in such a way that x is an obtuse angle. Laughing

eagle SG

thanks a lot for stating it so clearly. Wink

tanjun SG

No problem eagle. When writing all these replies, it is like recalling my past maths concepts that my teacher has taught me with her utmost effort. By the way, is my previous reply correct?

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