Calculating Boat Speed in 2-D Relative Motion

[BunDa4Th]

Suppose the river is moving east at 4.80 km/h and the boat is traveling Theta = 43.0° south of east with respect to the Earth. You will have to solve two equations and 2 unknowns.

(a) Find the speed of the boat with respect to the Earth. kh/m

(b) Find the speed of the boat with respect to the river if the boat's heading in the water is 60° south of east.
km/h

I am having trouble on where to start and what to do. I tried sqrt((4.80(cos43*))^2 + (4.80(sin43*))^2) but still give a wrong answer.

 

[mbrmbrg]

OK, let's look at part a.
I think:
Basic concept is $v_{BE}=v_{BR}+v_{RE}$
where V_BE= velocity of boat with respect to earth
v_BR is velocity of boat with respect to river
v_RE is velocity of river with respect to earth

Solve the above equation separately for x and y components, then use $v_{BE}=\sqrt{(v_{BEx})^2+(v_{BEy})^2}$

But my brain is fried. See if it makes sense, see if it works--good luck!

EDIT: never mind, the problem doesn't give you enough info to do all that. I guess you already know that, though... Sorry.

 

[BunDa4Th]

Yea, I tried doing what you said above but I did not have enough info to do it. I tried randomly doing things and closest I got the the correct answer was 14.8 but not sure how I did that.

 

[BunDa4Th]

Anyone? I been stuck on this question for the last 2 hour.

 

[mbrmbrg]

OK, take 2.

Those two angles are not two separate situations. Its the same exact boat at the same exact time doing the same exact thing. the 48 degree angle is with respect to the shore, and the 60 degree angle is with respect to... to... ah, hell, I don't know. And notice there's no handy right triangle to play with, you might have to draw another line to get one...

 

[BunDa4Th]

Okay, I tried using cos and sin in anyway I can but it seem to be no use
since that way i won't really learn anything from it.

 

[BunDa4Th]

Well, I'm a little bum out right now not able to get anywhere with this problem so maybe someone can shine some light here and help me out with this instead:

A daredevil decides to jump a canyon. Its walls are equally high and 12 m apart. He takes off by driving a motorcycle up a short ramp sloped at an angle of 14°. What minimum speed must he have in order to clear the canyon? m/s

I am not sure where to start with this problem. Any pointers would be great.

 

[BunDa4Th]

okay for the above problem this is what I did and still got the wrong answer.

V_oy = V_osin14*
V_ox = V_ocos14*

V_osin14* = (9.8)(1/2V_ocos14*)
2V_o^2(sin14)(cos14) = 9.8
V_o^2 = 9.8/(2(sin14)(cos14))
V_o = 4.54

which is incorrect when i plug in V_ox = 4.54cos14*

then i tried using this formula when I got deltay (which was probably wrong)
V_o = (sqrt(2G(Deltay))/(sin14) and i got 101 m/s which is also incorrect.

 

[6Stang7]

okay for the above problem this is what I did and still got the wrong answer.

V_oy = V_osin14*
V_ox = V_ocos14*

V_osin14* = (9.8)(1/2V_ocos14*)
2V_o^2(sin14)(cos14) = 9.8
V_o^2 = 9.8/(2(sin14)(cos14))
V_o = 4.54

which is incorrect when i plug in V_ox = 4.54cos14*

then i tried using this formula when I got deltay (which was probably wrong)
V_o = (sqrt(2G(Deltay))/(sin14) and i got 101 m/s which is also incorrect.

you're off witht he right foot in breaking the velocity up, but after that i am not sure what you are trying to do. it looks like you have a key idea here that the x and y parts are independent of one another, but that one will let you solve for the other. trying using the position function to solve this one.

first, we know the V_x=(V_0)cos14. so you have the velocity function and you need the position function. how would you go about getting that? once you have the position function (with respect to time), what would the position function be for the y direction and how would you relate it to the x direction?

 

[6Stang7]

Suppose the river is moving east at 4.80 km/h and the boat is traveling Theta = 43.0° south of east with respect to the Earth. You will have to solve two equations and 2 unknowns.

(a) Find the speed of the boat with respect to the Earth. kh/m

(b) Find the speed of the boat with respect to the river if the boat's heading in the water is 60° south of east.
km/h

I am having trouble on where to start and what to do. I tried sqrt((4.80(cos43*))^2 + (4.80(sin43*))^2) but still give a wrong answer.

you give no velocity for the boat itself, just the rivers.

 

[radou]

Okay, first let's write what we already know, so $$\vec{v}_{BE}=\vec{v}_{BR}+\vec{v}_{RE}$$. $$\vec{v}_{RE}=4.8\vec{i}, \vec{v}_{BR}=v_{BR}cos60\vec{i}+v_{BR}sin60\vec{j}, \vec{v}_{BE}=v_{BE}cos43\vec{i}+v_{BE}sin43\vec{j}$$. Now all we have to do is plug these equations into the equation $$\vec{v}_{BE}=\vec{v}_{BR}+\vec{v}_{RE}$$, from which we can easily retrieve two scalar equations with two unknowns (one for the i direction, and one for the j direction).

 

[BunDa4Th]

Okay, I am completely lost. It seem that the teacher did not go over this over the lecture last week. I tried looking up in my notes and pluging the equations into the equation but do not get what to do. I seriously need a new teacher...

 

[6Stang7]

let's see if this helps.

we know that the x component of the velocity is V_x=V_0cos14.
if we integrate this equation with respect to time we get X=V_0cos14(t).

now, what would be the velocity function equation for the y direction? what would the position function be?

once you have that, how could you relate the y-position function to the x-position function?

if you work with me on this you end up getting some very helpful equation for projectile motion.

 

[BunDa4Th]

if X = V_0cos14(t) i would plug that in Deltay = V_0y(t) - 1/2(g)(t)^2

which is t = x/(V_0cos14) which is
deltay = V_0y(x/(V_0cos14) - 1/2(-9.8)((x/(V_0cos14)^2

Sorry but physics is a really hard subject for me to understand but i know with help I will eventually get it.

 

[6Stang7]

you're doing perfect!
now, what is V_0y equal to? when you place that into the equal and simplfy, what do you get?

 

[BunDa4Th]

is V_0y = 0?

deltay = - 1/2(-9.8)((x/V_0cos14)^2 does x = 12?
deltay = -1/2(-9.8)(12.37/V_0^2)

im not sure if i should go further since i think i made a mistake somewhere

 

[6Stang7]

well, v_0y != 0 though. what is the vector component of the y velocity? you're getting really close to the answer though.

 

[BunDa4Th]

deltay = 60.61/V_0^2

V_0^2deltay = 60.61

deltay = 60.61V_0^2 if that is it do i sqrt to get rid of the V_0^2 or did i made a mistake.

 

[6Stang7]

we are not at the point for solving anything just yet; we are still doing equation manipulation here.

you said

if X = V_0cos14(t) i would plug that in Deltay = V_0y(t) - 1/2(g)(t)^2

which is t = x/(V_0cos14) which is
deltay = V_0y(x/(V_0cos14) - 1/2(-9.8)((x/(V_0cos14)^2

which is right and exactly what direction you need to be heading in.
if V_0=V, then what are the vector components (V_0y and V_0x)?

how can you use them in y=V_0y(x/(V_0cos14) - 1/2(-9.8)((x/(V_0cos14)^2?

 

[BunDa4Th]

okay V_0y would be 0 since there is no initial velocity for it.

so would y = V_0sin14

making it V_0sin14 = -1/2(-9.8)(x/(V_0cos14)^2

 

[6Stang7]

V_0y is the initial velocity in the y direction. we take it that when the motorcycle hits the ramp (and it at a angle of 14 degrees in this case). so the initial velocity, V_0, is at an anlge of 14 degress.
this makes V_0y what?

 

[BunDa4Th]

V_0y = V_0sin14

 

[6Stang7]

exactly. now, knowing that what does the equation we have been manipulating become?

 

[BunDa4Th]

y=V_0sin14(x/(V_0cos14) - 1/2(-9.8)((x/(V_0cos14)^2

btw, thanks for taking the time to help me out and explaining things.

 

[6Stang7]

it's my pleasure.

ok, so we have y=V_0sin14(x/(V_0cos14) - 1/2(-9.8)((x/(V_0cos14)^2. we can simplfy this equation by changing the V_0sin14(x/(V_0cos14) into V_0tan14(x) giving us y=V_0tan14(x) - (1/2(-9.8))/((V_0cos14)^2)(x^2)

in this case, V_0tan14 and (1/2(-9.8))/((V_0cos14)^2) are both constants, so for simplifcation let's say that:
V_0tan14=C1 and (1/2(-9.8))/((V_0cos14)^2)=C2

so now we have y=C1(x)-C2(x^2)

now, x is how far the object went in the x direction correct? so x can also be called the range
so, y=C1(R)-C2(R^2)

does this make since?

 

[BunDa4Th]

okay how does V_0sin14(x/V_0cos14) to V_0tan14(x)?

I see how you simplify it so the equation can be read easier but I am not really sure how i would solve y=C1(R)-C2(R^2)

 

[6Stang7]

ok, well V_0sin14(x/V_0cos14)=(V_0sin14(x)/V_0cos14)=(V_0sin14/V_0cos14)(x)=V_0tan14(x)

does that make since?

 

[BunDa4Th]

Oh okay I get it now. I see what you did now.

 

[6Stang7]

ok, so we have y=C1(R)-C2(R^2) where R is the range (or distance the object went in x direction). so, when does a projectial reach its max range?

 

[BunDa4Th]

what do you mean by max range? is it the distance that was given which is 12?

or

y/C1(R)-C2 = R^2?

 

[6Stang7]

well, in this problem the smallest max range needed is 12m, but that is for later.

what i mean is when the object has reached its max range it has travled the max distance in the x direction. if you threw a ball, the max range would be the distance from you to where the ball landed (in the x direction).

so when the object has reached its max range, what else do we know? what is the object height when it has reached the max range?

 

[BunDa4Th]

okay, now I am at a lost. I understand what you mean but I don't understand how I would set the equation to find the max range. and to find the max height i believe I would use V_y^2 = V_0y^2 - 2G(deltay)

 

[6Stang7]

ok, let's see if we can make this a little easier.
we have y=C1(R)-C2(R^2).
when the object has reached its max range, it has finished and is at the end of its trajectory. this means that the height is 0, or that y=0.

so if y=0 for max range, then y=0=C1(R)-C2(R^2).
so we now know that the max range of a projectial is C1(R)-C2(R^2)=0.
now solve this equation for R and what do you get?

 

[BunDa4Th]

C1r = C2r^2

C1r/r = C2r^2/r = C1 = C2r

R = C1/c2

 

[6Stang7]

BINGO!
now, re-write that equation by replacing C1 and C2 with what they are.