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?
To show that the motion is consistent with constant acceleration, you need to demonstrate that the differences in position and time between each step are related to each other by constant acceleration.
Here's how you can approach this problem:
Step 1: Find the velocity at point A
The velocity at point A can be calculated using the displacement and time of the first step. The displacement vector between O and A is given as OA = -2i - 4j, which means the object has moved 2 units in the negative x-direction and 4 units in the negative y-direction. The time taken is 2 seconds.
Using the formula: velocity = displacement / time,
the velocity at point A can be calculated as VA = (-2i - 4j) / 2 = -i - 2j.
Step 2: Find the acceleration from O to A
To find the acceleration from O to A, you need to consider the initial velocity (u) and the final velocity (VA), as well as the time taken (2 seconds).
Using the formula: acceleration = (final velocity - initial velocity) / time,
the acceleration from O to A can be calculated as A1 = (VA - u) / 2.
Step 3: Find the velocity at point B
The velocity at point B can be calculated using the displacement and time of the second step. The displacement vector between A and B is given as AB = 10i - 40j, which means the object has moved 10 units in the positive x-direction and 40 units in the negative y-direction. The time taken is 3 seconds.
Using the formula: velocity = displacement / time,
the velocity at point B can be calculated as VB = (10i - 40j) / 3.
Step 4: Find the acceleration from A to B
To find the acceleration from A to B, you need to consider the initial velocity (VA) and the final velocity (VB), as well as the time taken (3 seconds).
Using the formula: acceleration = (final velocity - initial velocity) / time,
the acceleration from A to B can be calculated as A2 = (VB - VA) / 3.
Step 5: Check if the accelerations are the same
Compare the calculated accelerations A1 and A2. If they are equal, then the motion is consistent with constant acceleration.
If A1 = A2, you have successfully shown that the motion is consistent with constant acceleration.
Step 6: Find the common acceleration and initial velocity
If A1 = A2, then you can solve for the common acceleration (a) and the initial velocity (u).
You will have two equations:
A1 = a
A2 = a
Solve these equations simultaneously to find the values of a and u.
I hope this explanation helps you understand how to approach this problem. Good luck solving it!