Thread: Intercepting vectors

Rower A is stationary at the point A adjacent to the river bank when he spots an object floating down the river. He estimates that the object is 50m away and heading south at a speed of 3m/s (the same speed as the current) see diagram below.

Assuming rower A can sustain a rowing speed of 4m/s

1 Draw a diagram of the velocity vectors involved if rower a is to intercept the object

2 calcuate the direction with respect to the river bank he needs to row (in degrees) in order to achieve his goal.

Now part 1 of the question i know how to do, but with part 2 i'm stuck, not really sure where to begin. If someone could just give a few hints then hopefully I can take it from there.

Many thanks

let be the desired angle for rower A's velocity w/r to the river bank.

using the diagram as reference, the rower's desired horizontal displacement is meters.

rower's velocity in the horizontal direction would be ...


so ...






in the vertical direction relative to the shore, the object's distance traveled + the rower's distance traveled must sum to be meters ...

relative to the shore in the vertical direction ...
object's speed = 3 m/s
rower's speed = m/s





all that is left is to sub in the first expression for t into the above equation and solve for ... which, btw, is really easy in this particular case.

Many thanks, I'll not pretend to have totally grapsed everything you've said, but i am working my way through bit by bit. Its coming abiet a bit slowly!

thanks again

Hello, ally79!

Rower is stationary at the point adjacent to the river bank
when he spots an object floating down the river.
He estimates that the object is 50 m away and heading south at a speed of 3 m/s.

Assuming rower can sustain a rowing speed of 4 m/s,

1) Draw a diagram of the velocity vectors involved if rower is to intercept the object.

2) Calculate the direction with respect to the river bank he needs to row
(in degrees) in order to achieve his goal.

Code:

      N           * B
      |          *|
      |         * |
      |        *  |
      |    50 *20°| 3t
      |      *    |
      |     *     |
      |    *      |
      |20°*       * C
      |  *     *
      | *θ  * 4t
      |* *
    A *
      |

We are given: .
. . Note that:

In seconds, has moved m to point

In seconds, has moved m to point

Let

We want


Using the Law of Sines: .

Hence: .


Therefore: .


Edit: skeeter pointed out an error . . . *blush*
I forgot: the boat is also affected by the current.

I'll have to work on a corrected version.

Sorry . . .
.

Skeeter,
I follwed through what you said and I was able to follow it and understand it, Hower when substuting in t and rearranging to solve for theta I ended up with 50 Sin (20)/50cos (20) = 4 Sin theta/4 Cos Theta

Which i cancelled down to Tan 20 = Tan Theta Which gives me the original 20 degree angle.

If anyone can point out where I went wrong i'd be grateful!

ironic, isn't it? but correct.

is that a further 20 degrees on top of the 20 degrees already?

How can the angle of interception be 20 degrees when the initial angle is 20 degrees? Surely as the object is floating down stream the angle will be increasing every second?

Can someone please explain this to me? I apologise if i'm missing something obvious

Originally Posted by ally79

is that a further 20 degrees on top of the 20 degrees already?

no ... the 20 degrees relative to the shore is what the boat must steer to intercept the object.