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Kinematics Physics Help

  • #1
Hi, I'm new to this physics forum. I need guidance/direction on how to tackle the following problems: (any help would be appreciated at this time)

1) A race car starts from rest on a circular track. The car increases its speed at a constant rate at as it goes once around the track. Find the angle that the total acceleration of the car makes-with the radius connecting the center of the track and the car-at the moment the car completes the circle.


2) A Coast Guard cutter detects an unidentified ship at a distance of 20.0 km in the direction 15.0° east of north. The ship is traveling at 26.0 km/h on a course at 40.0° east of north. The Coast Guard wishes to send a speedboat to intercept the vessel and investigate it. If the speedboat travels 50.0 km/h, in what direction should it head? Express the direction as a compass bearing with respect to due north.

Where do I start??!?!

Thanks =)
 

Answers and Replies

  • #2
SGT
1) The acceleration has two components: a constant tangential acceleration [tex]\vec a_T[/tex], responsible for the increase in tangential velocity and a radial acceleration [tex]\vec a_R[/tex] responsible for the circular movement. The total acceleration is the resultant of [tex]\vec a_T[/tex] and [tex]\vec a_R[/tex].

2) You have a triangle OAB. O(0,0) is the position of the cutter and of the speedboat. A(ax,ay) = (10sin15,10cos15) is the position of the ship. B(bx,by) is the intersection of the two trajectories.
[tex]bx = ax + 26\cdot sin40 \cdot T= 0 + 50 \cdot sinC \cdot T[/tex]
[tex]by = ay + 26\cdot cos40 \cdot T= 0 + 50 \cdot cosC \cdot T[/tex]
Where C is the speedboat course and T is the time for interception.
 
  • #3
further explanation...plz?

SGT said:
1) The acceleration has two components: a constant tangential acceleration [tex]\vec a_T[/tex], responsible for the increase in tangential velocity and a radial acceleration [tex]\vec a_R[/tex] responsible for the circular movement. The total acceleration is the resultant of [tex]\vec a_T[/tex] and [tex]\vec a_R[/tex].

2) You have a triangle OAB. O(0,0) is the position of the cutter and of the speedboat. A(ax,ay) = (10sin15,10cos15) is the position of the ship. B(bx,by) is the intersection of the two trajectories.
[tex]bx = ax + 26\cdot sin40 \cdot T= 0 + 50 \cdot sinC \cdot T[/tex]
[tex]by = ay + 26\cdot cos40 \cdot T= 0 + 50 \cdot cosC \cdot T[/tex]
Where C is the speedboat course and T is the time for interception.
For the 1) do u get an answer of 1/4pi=tan theta which = 4.55 degrees? 2) where did u get 10 from in the (10sin15,10cos15)?? and what equation is that.. your using for bx and by? What is the final answer for #2)?

sorry for all the trouble, i'm very confused when it comes to these types of physics problems.
 
  • #4
SGT
physixnot4me said:
For the 1) do u get an answer of 1/4pi=tan theta which = 4.55 degrees? 2) where did u get 10 from in the (10sin15,10cos15)?? and what equation is that.. your using for bx and by? What is the final answer for #2)?

sorry for all the trouble, i'm very confused when it comes to these types of physics problems.
10 is the distance: 10 km. If you prefer to use meters, replace it by 10000. In the same way, I used 26 and 50 for the speeds in km/h. If you want to use meters, you should use 26000/3600 m/s and 50000/3600 m/s.
 
  • #5
SGT said:
10 is the distance: 10 km. If you prefer to use meters, replace it by 10000. In the same way, I used 26 and 50 for the speeds in km/h. If you want to use meters, you should use 26000/3600 m/s and 50000/3600 m/s.
Sorry, I still don't see where your getting 10km from. Did you figure that out on your own, using the values already given to you in the question?.. Given was 26km, 20km/h and 50km/hr. Is there another way of solving this problem without using components? Since this problem can only be solved, by gettin the diagram right to begin with, I was just wondering if there was an alternate solution which may be easier to visualize using the diagram I've drawn for myself.
 
  • #6
SGT
physixnot4me said:
Sorry, I still don't see where your getting 10km from. Did you figure that out on your own, using the values already given to you in the question?.. Given was 26km, 20km/h and 50km/hr. Is there another way of solving this problem without using components? Since this problem can only be solved, by gettin the diagram right to begin with, I was just wondering if there was an alternate solution which may be easier to visualize using the diagram I've drawn for myself.
My mistake. Where I wrote 10, read 20. I don't think you could solve the problem without using components. Since you have two unknowns: the course of the speedboat and the time for interception, you need two equations. The easiest way to obtain them is using components.
 
  • #7
SGT said:
My mistake. Where I wrote 10, read 20. I don't think you could solve the problem without using components. Since you have two unknowns: the course of the speedboat and the time for interception, you need two equations. The easiest way to obtain them is using components.
I was also confused about why you switched the sin and cos. I always learned that x= cos theta and y= sin theta... but in the components you gave, you have sin in the 'x' and cos in the 'y'... ???
 
  • #8
physixnot4me said:
I was also confused about why you switched the sin and cos. I always learned that x= cos theta and y= sin theta... but in the components you gave, you have sin in the 'x' and cos in the 'y'... ???
Also, in the 2 eqns you have for bx and by... Do I solve for T? in Bx, then sub T back into By.. to solve for C?
 
  • #9
SGT
physixnot4me said:
I was also confused about why you switched the sin and cos. I always learned that x= cos theta and y= sin theta... but in the components you gave, you have sin in the 'x' and cos in the 'y'... ???
Because in geographical coordinates zero degree points North and the angles grow clockwise, while in the trigonometrical circle zero degree points to the right and the angles grow counterclockwise.
 
  • #10
SGT
physixnot4me said:
Also, in the 2 eqns you have for bx and by... Do I solve for T? in Bx, then sub T back into By.. to solve for C?
Yes, that is what you must do.
 
  • #11
Where am i going wrong ?

[tex]bx = ax + 26\cdot sin40 \cdot T= 0 + 50 \cdot sinC \cdot T[/tex]

If u were to rearrange this for T:

[tex]T=ax / 50 \cdot sinC - 26\cdot sin40 [/tex]

This is how I have isolated for 'T'... and when i sub this into the equation you gave:

[tex]by = ay + 26\cdot cos40 \cdot T= 0 + 50 \cdot cosC \cdot T[/tex]

I have no idea where to begin when it comes to cancelling out... since there is now sin's introduced into the by equation(that contains all cos's)

HELP?? How would you even isolate for 'C' :bugeye:
 
  • #12
SGT
physixnot4me said:
[tex]bx = ax + 26\cdot sin40 \cdot T= 0 + 50 \cdot sinC \cdot T[/tex]

If u were to rearrange this for T:

[tex]T=ax / 50 \cdot sinC - 26\cdot sin40 [/tex]

This is how I have isolated for 'T'... and when i sub this into the equation you gave:

[tex]by = ay + 26\cdot cos40 \cdot T= 0 + 50 \cdot cosC \cdot T[/tex]

I have no idea where to begin when it comes to cancelling out... since there is now sin's introduced into the by equation(that contains all cos's)

HELP?? How would you even isolate for 'C' :bugeye:
Remember that [tex]26\cdot sin40 [/tex] and [tex]26\cdot cos40 [/tex] are constants. Also, you can make [tex]50 = k \cdot sin \phi [/tex] and use the formulas for the products of sines and cosines.
Alternatively, you can isolate [tex]sinC[/tex] in the first equation and [tex]cosC[/tex] in the second. The quotient of both is [tex]tanC[/tex].
 

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