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# Advanced Math/2 dimensional motion with constant acceleration

Question
QUESTION: This is a question from Mechanics 1.
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An object is moving in a plane.  At time t =0, it is at the origin O and moving with velocity u.  After 2 seconds it is at A where OA(vector)= -2i -4j.  After a further 3 second it is at B where AB(vector) is 10i - 40j.

Show that this is consistent with constant acceleration a,  Find a and u.
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Do I have to find the velocity at A and then use it to find the velocity at B?  I've tried this but I'm not getting very far.  Any clues would be most appreciated.  I'm keen to know how to solve it more than just the solution itself so any commentary would be of great help to me.

Thanks

Maggie

ANSWER: In this problem, we need to calculate the velocities and accelerations as derivatives of position with respect to time. This is done using the differences of appropriate values at the specified times. The relationships from basic physics are (using calculus notation)

velocity =  d(position)/dt =       ∆x/∆t   in meters/second (m/s)
acceleration = d(velocity)/dt =  ∆u/∆t   in (m/s^2)

Since the quantities are vectors, we can simplify things by solving the problem in its x and y components separately. Using

t1 = 0,   x1 = 0,     y1 = 0
t2 = 2,   x2 = -2,   y2 = -4
t3 = 3,   x3 = 10,   y3 = -40

For the 1st time segment from 0 to 2 seconds, the velocities are

ux1 = ∆x/∆t  =  (x2-x1)/(t2-t1)  =  (-2-0)/(2-0)  =  -1 m/s
uy1 = ∆y/∆t  =  (y2-y1)/(t2-t2)   =  (-4-0)/(2-0)  =  -2 m/s

For the 2nd segment

ux2 =  (x3-x2)/(t3-t2)  =  (10-(-2))/(3-2)  =  12 m/s
uy2 =  (y3-y2)/(t3-t2)   =  (-40-(-4))/(3-2)  =  -36 m/s

Now we can calculate the accelerations; the time interval for ∆t will be the difference between the centers of each segment

∆t  =  (t3+t2)/2 - (t2+t1)/2 = (t3-t1)/2 = 3/2

ax  =  ∆ux/∆t  =  (ux2-ux1)/(3/2)  = (12-(-1))･(2/3)    =  -26/3 m/s^2
ay  =  ∆uy/∆t  =  (uy2-uy1)/(3/2)   = (-36-(-2)･(2/3)    =  -34/3 ms^2.

The total vector acceleration a = (ax, ay).

To find the initial velocities ux0 and uy0, use the definition of acceleration (which is constant) in the 1st segment

(ux1-ux0)/(t2-t1)  = ax  ->  ux0 = ux1 - ax･(t2-t1) = -1 - (-26/3)･2 =  16.3 m/s
(uy1-uy0)/(t2-t1)  = ay  ->   uy0 = uy1 - ay･(t2-t1) =  -2 - (-34/3)･2 =  24.7 m/s.

Hope this helps. It really is just a bunch of bookkeeping once you have the relationships between x, u and a.

Randy

---------- FOLLOW-UP ----------

QUESTION: Oh dear

It's obvious that I don't know enough.  For the second segment, why are you dividing by (3-2).  I thought it would be just 3 because it took a further 3 seconds to get there.

I'm also struggling to see where "Show that this is consistent with constant acceleration a" happens.  I thought that this was asking us to show that the acceleration was the same in both segments of the journey and we somehow have to use "u" as the velocity at the origin.

Oh well, although I don't totally get what you've done it's given me loads to think about.  The following bit is completely new to me.

"Now we can calculate the accelerations; the time interval for ∆t will be the difference between the centers of each segment

∆t  =  (t3+t2)/2 - (t2+t1)/2 = (t3-t1)/2 = 3/2

ax  =  ∆ux/∆t  =  (ux2-ux1)/(3/2)  = (12-(-1))･(2/3)    =  -26/3 m/s^2
ay  =  ∆uy/∆t  =  (uy2-uy1)/(3/2)   = (-36-(-2)･(2/3)    =  -34/3 ms^2."

Thanks very much for taking all that time and your commentary was just what I needed.

Maggie

Maggie, you're right about the second segment being 3 secs vs 1. Sorry for the confusion. The "consistent with constant acceleration" part is more subtle and it is a little puzzling the way it is worded.

As described in my first response, velocity over a time segment is calculated by dividing the separation in distance by the change in time, a so-called finite difference, which approximates the time derivative. It seems like you understand this part. Now, the acceleration is calculated as the finite difference in velocity, which means we need the velocity at 2 points. The 3 points in the problem mark the position and time of the object and we needed all three to calculate 2 velocities. We then use these 2 values, with an appropriate time interval, to calculate the acceleration. In short, using finite differences, the 3 position and time points give 2 velocity estimates and 1 acceleration.

The acceleration is assumed constant which enabled the velocity at the origin to be calculated. Since, at first glance, there is only 1 acceleration that can be calculated, it doesn't seem possible to determine if it is constant. If, on the other hand, the trajectory of he object is sketched, it changes direction along the x-axis, which might suggest that there is a changing acceleration (i.e., force). However, decomposing the trajectory into its component parts along the x and y axes shows that a constant acceleration is consistent.

Thanks for your question and follow-up. It is refreshing to get a real question from someone instead of just a "cut and paste" homework problem. Good luck with your studies and please ask me any questions you come across.

Randy
Questioner's Rating
 Rating(1-10) Knowledgeability = 10 Clarity of Response = 10 Politeness = 10 Comment Thanks. That all makes sense to me now. You have explained about the acceleration - being able to calculate it from any two velocity points with a known time difference. I thought I'd have to do it from end points. Umm. Must try to remember this in the future. As I'm only on chapter 2 of a 6 chapter book and this is the beginners' book I can see there's a long way to go and potential to use that information again. Cheers. Maggie

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#### randy patton

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