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?
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.
To show that the motion is consistent with constant acceleration, you need to demonstrate that the velocity changes linearly with time. This means that the object's velocity increases or decreases at a constant rate throughout its motion.
To solve the problem, you can break it down into two steps:
Step 1: Determine the acceleration between O and A.
Step 2: Determine the acceleration between A and B.
Step 1: Determining the acceleration between O and A
To find the acceleration between O and A, you can use the formula:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity is u, the time is 2 seconds, and the final velocity is the velocity at point A.
The equation can be rewritten as:
-2i - 4j = u + a(2).
Since the acceleration is constant, you can assume the object started at rest. Hence, the initial velocity, u, is 0.
The equation becomes:
-2i - 4j = a(2).
This equation gives you the acceleration, a, between O and A.
Step 2: Determining the acceleration between A and B
Now, you need to determine the acceleration between A and B. Again, you can use the same formula:
v = u + at.
This time, the initial velocity is the velocity at A, and the final velocity is the velocity at B. The time is 3 seconds.
The equation can be rewritten as:
10i - 40j = (-2i - 4j) + a(3).
Simplifying this equation, you get:
a = (10i - 40j - (-2i - 4j))/(3).
This equation gives you the acceleration, a, between A and B.
Finally, to find the initial velocity, u, you can use the equation:
v = u + at,
where v is the final velocity at B, and t is the time taken to go from A to B.
The equation can be rewritten as:
10i - 40j = u + a(3).
Substituting the value of a that you found in the previous step, you can solve for u.
Once you have found the values of a and u, you have successfully shown that the motion is consistent with constant acceleration.
Note: It is important to carefully handle the vectors (i and j components) while solving these equations.
I hope this explanation helps you understand the steps to solve the problem. Good luck!