Whether the boy can catch the ship? relative velocity

In summary: Since all three angles are equal, and \alpha is less than \gamma, then the sum of the vectors is less than the original vector so the boy meets the boat at an angle of \gamma.
  • #1
jessicaw
56
0

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!
 

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  • #2


type the problem statement... your posted file's font is too small to read
 
  • #3


very urgent i have upload a clear file now thanks for answering
 
  • #4


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
 
  • #5


godtripp said:
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.
 
  • #6


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.
 
  • #7


godtripp said:
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.
 
  • #8


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

[tex]\bar{u}t=\bar{\mathrm{v}_1}t_1+\bar{\mathrm{v}_2}t_2[/tex] 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 [tex]\theta[/tex] with the river bank. and let
[tex]\alpha[/tex] be the angle between [tex]\bar{v_1}[/tex] and [tex]\bar{v_2}[/tex].

so [tex]\alpha[/tex] 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
[tex]ut[/tex] is [tex]\pi-\alpha[/tex]. angle in front of the side with length equal to
[tex]v_2 t_2[/tex] is [tex]\theta[/tex] and the angle in front of the side with length equal
to [tex]v_1 t_1 [/tex] is (from geometry) [tex]\alpha-\theta[/tex]. so law of sines gives

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

from here we get (since t=t1+t2)

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

and

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

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

[tex]u=\frac{v_1 v_2 \sin \alpha}{v_2 \sin(\alpha-\theta)+v_1 \sin \theta}[/tex]

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

[tex]\cos \alpha=\frac{v_2}{v_1}[/tex]

so the maximum value of u becomes

[tex]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 } [/tex]

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

1. What is relative velocity?

Relative velocity is the velocity of an object in relation to another object. It takes into account the motion of both objects and is measured in terms of distance traveled per unit of time.

2. How is relative velocity calculated?

Relative velocity is calculated by subtracting the velocity of one object from the velocity of the other object. This gives the relative velocity between the two objects.

3. How does relative velocity affect the boy's ability to catch the ship?

The relative velocity between the boy and the ship will determine whether or not the boy is able to catch the ship. If the boy's relative velocity is greater than the ship's, he will be able to catch it. If the ship's relative velocity is greater, the boy will not be able to catch it.

4. Can relative velocity change?

Yes, relative velocity can change depending on the motion of the objects. If either object changes its velocity, the relative velocity between them will also change.

5. How does direction affect relative velocity?

Direction plays a crucial role in calculating relative velocity. If both objects are moving in the same direction, their relative velocity will be different than if they are moving in opposite directions. This is because the direction of motion affects the overall velocity of the objects.

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