# Intercepting vectors

• Sep 19th 2008, 11:27 AM
ally79
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
• Sep 19th 2008, 01:04 PM
skeeter
let $\displaystyle \theta$ 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 $\displaystyle 50\sin(20)$ meters.

rower's velocity in the horizontal direction would be ...
$\displaystyle v_x = 4\sin{\theta}$

so ...

$\displaystyle 4\sin{\theta} \cdot t = 50\sin(20)$

$\displaystyle t = \frac{50\sin(20)}{4\sin{\theta}}$

in the vertical direction relative to the shore, the object's distance traveled + the rower's distance traveled must sum to be $\displaystyle 50\cos(20)$ meters ...

relative to the shore in the vertical direction ...
object's speed = 3 m/s
rower's speed = $\displaystyle 4\cos{\theta} - 3$ m/s

$\displaystyle 3t + (4\cos{\theta}-3)t = 50\cos(20)$

$\displaystyle 4\cos{\theta} \cdot t = 50\cos(20)$

all that is left is to sub in the first expression for t into the above equation and solve for $\displaystyle \theta$ ... which, btw, is really easy in this particular case.
• Sep 19th 2008, 01:39 PM
ally79
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
• Sep 19th 2008, 01:44 PM
Soroban
Hello, ally79!

Quote:

Rower $\displaystyle A$ is stationary at the point $\displaystyle A$ 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 $\displaystyle A$ can sustain a rowing speed of 4 m/s,

1) Draw a diagram of the velocity vectors involved if rower $\displaystyle A$ 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: .$\displaystyle \angle NAB = 20^o,\;AB = 50$
. . Note that: $\displaystyle \angle ABC = 20^o$

In $\displaystyle t$ seconds, $\displaystyle A$ has moved $\displaystyle 4t$ m to point $\displaystyle C.$

In $\displaystyle t$ seconds, $\displaystyle B$ has moved $\displaystyle 3t$ m to point $\displaystyle C.$

Let $\displaystyle \theta = \angle BAC$

We want $\displaystyle \angle NAC.$

Using the Law of Sines: .$\displaystyle \frac{\sin\theta}{3t} \:=\:\frac{\sin20^o}{4t} \quad\Rightarrow\quad \sin\theta \:=\:\frac{3\sin20^o}{4} \:=\:0.256515107$

Hence: .$\displaystyle \theta \:=\:14.86338095^o \:\approx\:14.86^o$

Therefore: .$\displaystyle \angle NAC \;=\;14.86^o + 20^o \;=\;\boxed{34.86^o}$

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 . . .
.
• Sep 23rd 2008, 12:53 PM
ally79
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!
• Sep 23rd 2008, 06:04 PM
skeeter
ironic, isn't it? but correct. (Nod)
• Sep 23rd 2008, 08:26 PM
ally79
is that a further 20 degrees on top of the 20 degrees already?
• Sep 24th 2008, 10:12 AM
ally79
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
• Sep 24th 2008, 02:13 PM
skeeter
Quote:

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.