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Calculate the velocity vector as a function of time

Hi,

Ok I have a bit of difficulty workin this Physics Mechanics problem.

Ok here is the problem....

The coordinates of a plane flying in the xy-plane are given as functions of time by:

X = 2.0m - at
Y = bt^2

where a = 3.0 m/s and b = 2.5 m/s^2

A) Calculate the velocity vector as a function of time.
B) Calculate the acceleration vector as a function of time.
C) Calculate the magnitude and direction of the bird's velocity at t=2.9s.
D) Calculate the magnitude and direction of the bird's acceleration a t=2.9s.

Ok here's what i've accomplished so far...

Code:
                                NOTE:     
                    /^          a(x) = a subscript x
    y|             / |          a(y) = a subscript y
     |            /  |             i = unit vector component, x-axis
     |         a /   |             j = unit vector component, y-axis
     |          /    |a(y)j        
     |         /     |
     |        /θ     |
bt^2 -       *------->
     |         a(x)i
     |
     --------|---------------
          2.0 - at          x
Now I have these formulas.
Code:
    a = √((a(x))^2 + (a(y))^2)
tan θ = a(y)/a(x)
 a(x) = a cos θ   
 a(y) = a sin θ  
    a = a(x)i + a(y)j
Ok, now I believe you find a(x) and a(y) by taking the derivatives of bt^2 and 2.0 - at, thus getting 2bt and -a.

But I don't quite understand what the question (part A) is asking, when they want the velocity vector as a function of time.

I'm guessing its this: v(t) = -3.0i + 5.0tj

if this is so then the acceleration vector (part B) is expressed as:
a(t) = 5.0j

So based on this i'm guessing part C is:

v(2.9s) = 5.83 m/s

but i'm not sure if this is correct either since part c asks for 'magnitude and direction', I feel like i'm missing something to my answer...

Lastly following all the above part D should be:

a(2.9s) ≈ 5.0 m/s^2

So....I hope I can get some insight in to this problem, I hope someone can help, and give me some guidance.

Thanks.
 

HallsofIvy

Science Advisor
Homework Helper
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Ok, now I believe you find a(x) and a(y) by taking the derivatives of bt^2 and 2.0 - at, thus getting 2bt and -a.

But I don't quite understand what the question (part A) is asking, when they want the velocity vector as a function of time.
Your first statement is incorrect. Acceleration is the derivative of velocity. What you are given is the "position" vector.

You know that X= 2.0 - 3.0t and Y = 2.5t^2 which means that the position vector is (2.0- 3.0t)i+ (2.5t^2)j. Differentiate that to get the velocity vector: velocity vector = -3.0 i+ 5.0t j, exactly what you have. The derivative of that is acceleration vector = 5.0j, again, the answer you have.

C) Calculate the magnitude and direction of the bird's velocity at t=2.9s.
How did the plane turn into a bird?

At t=2.9, the velocity vector is -3.0i+ 5.0(2.9)j= -3.0i+ 14.5j. The simplest way to calculate the "magnitude" (i.e. speed) is to use the Pythagorean theorem. The length of the vector is √((-3)2+(14.5)2)= 14.8 m/s. Since you don't say how you arrived at "5.83 m/s", I have no idea how you got that. The angle is, as your formula says, tan-1(Vy/Vx)= tan-1(14.5/-3)= tan-1(-4.833)= -0.08, only very very slightly less below the x axis.

D) Calculate the magnitude and direction of the bird's acceleration a t=2.9s.
The acceleration vector at any time is 5.0j which has magnitude 5 m/s2, which you have (and, by the way, that is exact, not approximate) and direction (which you left out) straight up the y axis.
 
Your first statement is incorrect. Acceleration is the derivative of velocity. What you are given is the "position" vector.
heh...Thanks for clearin that up, I kinda knew that but now it makes much more sense.

Ok since i'm given, X= 2.0 - 3.0t and Y = 2.5t^2, this is the position vector which is expressed as.

x(t) = (2.0- 3.0t)i + (2.5t^2)j

then: x'(t) = v(t)

v(t) = -3.0i + 5.0tj

then: v'(t) = a(t)

a(t) = 5.0j

How did the plane turn into a bird?
woops...yea that was a typo, its still a plane :)

And I guess it would have made more sense if I drew my original diagram as this:
Code:
                                NOTE:     
                    /^          v(x) = v subscript x
    y|             / |          v(y) = v subscript y
     |            /  |             i = unit vector component, x-axis
     |         v /   |             j = unit vector component, y-axis
     |          /    |v(y)j        
     |         /     |
     |        /?     |
bt^2 -       *------->
     |         v(x)i
     |
     --------|---------------
          2.0 - at          x
I only used 'a' as this was the notation used in my textbook I knew it was velocity already.

At t=2.9, the velocity vector is -3.0i+ 5.0(2.9)j= -3.0i+ 14.5j. The simplest way to calculate the "magnitude" (i.e. speed) is to use the Pythagorean theorem.
hey thanks for clarifying what magnitude is.

The length of the vector is √((-3)2+(14.5)2)= 14.8 m/s. Since you don't say how you arrived at "5.83 m/s", I have no idea how you got that. The angle is, as your formula says, tan-1(Vy/Vx)= tan-1(14.5/-3)= tan-1(-4.833)= -0.08, only very very slightly less below the x axis.
yea...it was just a careless error on my part I calculated 5.83 m/s by √(-3)^2 + (2.9)^2 instead of √(-3)^2 + (14.5)^2 = 14.8 m/s.

The acceleration vector at any time is 5.0j which has magnitude 5 m/s2, which you have (and, by the way, that is exact, not approximate) and direction (which you left out) straight up the y axis.
oh yea thanks, so then Part D is:

Acceleration:
direction = 5.0j
magnitude = 5 m/s^2

Ok thanks again, I have a way better grasp on this mechanics stuff now, heh.
 
Last edited:

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