Whether the boy can catch the ship? relative velocity

[jessicaw]

Homework Statement

see upload file

Homework Equations

The Attempt at a Solution

consider relative velocity and show that boy can go back to origin.
thanks!

 

[issacnewton]

type the problem statement... your posted file's font is too small to read

 

[jessicaw]

very urgent i have upload a clear file now thanks for answering

 

[godtripp]

Hmmm, since it doesn't say anything about the boy's direction when he starts swimming I assume they want you to consider a right triangle...of course this depends on your level of physics class also... if you find this is not the case then use the law of cosines to solve the problem .

Consider a right triangle with edge AB as the bank shore. edge AC as the path of the boat and edge BC as the path the boy swims to reach the boat
angle ABC = 90 deg
and angle BAC = 15 deg

now the problem becomes easy if you think of these edges as vectors So edge AC is the vector representing the boat's velocity. The edges CB and AB represent components of the boys velocity
then find the magnitude of the boy's velocity by sqrt ( CB^2 + AB^2 ) and violla ... v(max) = sqrt(20)

again... this is assuming the boys swims perpendicular to the shore.. if you're in a college course with a calculus prerequisite ... you might want to introduce some variable angle theta for angle ABC and do some calculus to find out which angle would give you a minimum time. (related rates) and then solve for the vector sum that way

 

[jessicaw]

Hmmm, since it doesn't say anything about the boy's direction when he starts swimming I assume they want you to consider a right triangle...of course this depends on your level of physics class also... if you find this is not the case then use the law of cosines to solve the problem .

Consider a right triangle with edge AB as the bank shore. edge AC as the path of the boat and edge BC as the path the boy swims to reach the boat
angle ABC = 90 deg
and angle BAC = 15 deg

now the problem becomes easy if you think of these edges as vectors So edge AC is the vector representing the boat's velocity. The edges CB and AB represent components of the boys velocity
then find the magnitude of the boy's velocity by sqrt ( CB^2 + AB^2 ) and violla ... v(max) = sqrt(20)

again... this is assuming the boys swims perpendicular to the shore.. if you're in a college course with a calculus prerequisite ... you might want to introduce some variable angle theta for angle ABC and do some calculus to find out which angle would give you a minimum time. (related rates) and then solve for the vector sum that way

the boys direction is actually arbitary.

 

[godtripp]

Direction should affect your time traveled by the boy, and thus your velocity of the boat.
Maybe you found a way the angles cancel out to an identity? I'm not sure I'm just considering the problem in my head.

 

[issacnewton]

Consider a right triangle with edge AB as the bank shore. edge AC as the path of the boat and edge BC as the path the boy swims to reach the boat
angle ABC = 90 deg
and angle BAC = 15 deg

now the problem becomes easy if you think of these edges as vectors So edge AC is the vector representing the boat's velocity. The edges CB and AB represent components of the boys velocity
then find the magnitude of the boy's velocity by sqrt ( CB^2 + AB^2 ) and violla ... v(max) = sqrt(20)

CB and AB do not represent the components of the boy's velocity. boy can run different
time on the ground and swim for different time. also you can't do velocity addition stuff here since all three velocities are relative to the ground only.

 

[issacnewton]

i think i got some way out of this.

lets assume that boy is able to meet the boat. i am assuming that the boy's running path
on the river bank is downwards if the boat is also going downwards. let boy run on the river bank for time t₁ and swim in the river for time t₂. since boy meets
the boat, we can add the displacement vectors

\bar{u}t=\bar{\mathrm{v}_1}t_1+\bar{\mathrm{v}_2}t_2 where t =t₁+t₂

is the total time of journey. during this time the boat also reaches that point of
meeting. let the boat make angle \theta with the river bank. and let
\alpha be the angle between \bar{v_1} and \bar{v_2}.

so \alpha is an external angle of the triangle formed by three displacement vectors.
we will now use law of sines here. the angle in front of the side with length equal to
ut is \pi-\alpha. angle in front of the side with length equal to
v_2 t_2 is \theta and the angle in front of the side with length equal
to v_1 t_1 is (from geometry) \alpha-\theta. so law of sines gives

\frac{v_2 t_2}{\sin \theta}=\frac{ut}{\sin \alpha}=\frac{v_1 t_1}{\sin(\alpha-\theta)}

from here we get (since t=t₁+t₂)

u=\frac{t_2}{(t_1+t_2)} \sin\alpha \, \frac{v_1t_1}{\sin(\alpha-\theta)} ...1)

and

t_2 =\frac{v_1}{v_2}\, \frac{\sin \theta}{\sin(\alpha-\theta)}\, t_1...2)

we can use eq 2 to eliminate t₁ and t₂ from eq 1. and we can simplify eq. 1 as

u=\frac{v_1 v_2 \sin \alpha}{v_2 \sin(\alpha-\theta)+v_1 \sin \theta}

since \theta is constant, and v₁ and v₂ are also constants, u is function of \alpha. we can now use methods of calculus to maximize u. when we equate the first derivative of u with respect to \alpha
to zero and evaluate the expression we get the condition,

\cos \alpha=\frac{v_2}{v_1}

so the maximum value of u becomes

u_{max}=\frac{v_1 v_2 \left[1-\frac{v_2^2}{v_1^2}\right]^{1/2} }{v_2 \left[1-\frac{v_2^2}{v_1^2}\right]^{1/2} \cos \theta - \frac{v_2^2}{v_1} \sin \theta + v_1 \sin \theta }

i hope my calculations and reasoning is correct. please check it for different special cases.