this is analogous to showing that
y(0) = 0
y(1) = 1
y(2) = 4
is consistent with constant acceleration
In that case, let acceleration be the constant 2a.
Then y'(x) = 2ax+b
y(x) = ax^2+bc+c
Use the three equations to show that
y=x^2 is consistent with the position values and constant acceleration.
Here, we have
r(0) = (0,0)
r(2) = (-2,-4)
r(5) = (10,-40)
a constant acceleration (2a,2b) and initial velocity (ux,uy) would give us
r'' = (2a,2b)
r' = (2at+ux,2bt+uy)
r = (at^2+ux t,bt^2+uy t)
So, plug in the r values to solve for a,b and u:
Initial velocity of (-3,2) and constant acceleration of (1,-2) will produce the position values given.
The problem I am trying to solve is from AQA Mechanics M1 and is as follows:
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 = -2i - 4j. After a further 3 seconds it is at B where AB = 10i - 40j. Show that this is consistent with constant acceleration a, Find a and u.
I don't understand what "Show that this is consistent with constant acceleration a"means, Do I have to find the acceleration between O and A and then between A and B and work out if they are the same? I've tried this with the initial velocity of u to find the acceleration for the first step and the velocity at the end of the first step then I tried to use the velocity as the start point for the second step with a different a hoping it would come out as the same value as a in the first step. The equations were enormous and I don't think this was the right approach. However, this question being unlike any example or other exercises in the chapter, I'm at a loss to know where to start.
Any clues?
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