Help with question

  • Thread starter Xtasy
  • Start date
  • #1
8
0
A bird flies in the xy-plane with a velocity vector given by v(vector) = (a-bt^2)(i-vector) + ct (j-vector) , with a= 2.4m/s, b=1.6m/s^3,c=4.0m/s^2. The positive y-direction is vertically upward. At t=0 the bird is at the origin.

Calculate the position vector of the bird as a function of time.
Express your answer in terms of a, b, and z. Write the vector r(t) in the form vector(r(t)x,r(t)y , , where the x and y components are separated by a comma.

Calculate the acceleration vector of the bird as a function of time. also in the form as stated above.

What is the bird's altitude (y-coordinate) as it flies over x=0 for the first time after t=0?
 

Answers and Replies

  • #2
Pyrrhus
Homework Helper
2,178
1
I see, good standard questions... :rolleyes:

Let's see how you are tackling them!
 
  • #3
8
0
Well, atiderviative will give me the position vector, but I don't know if I should plug in the values of a, b, etc. prior to me taking the antiderivative. Also I am not sure what form they are talking about.

For the acceleration the derivative is needed.

Not sure about the last question
 
  • #4
Pyrrhus
Homework Helper
2,178
1
a, b, c are constants.

For the last question look at the position function, what can you notice?
 
  • #5
8
0
I notice that x=0, so there is no horizontal motion. What do they mean by birds altitude as it flies over x=0 after t=0, meaning whent t=1. Also so since x=0 I only focus on the y componet of the position vector, which is (ct^2)/2?
 
  • #6
Pyrrhus
Homework Helper
2,178
1
Look at the acceleration vectorial function, it's x component is negative, so sooner or later the bird will pass the straight line x=0.
 

Related Threads on Help with question

  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
8
Views
5K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
821
  • Last Post
Replies
4
Views
2K
Top