An object is moving in a plane. At time t =0,it is at origin O, and moving with velocity u. After 2 seconds, it is at A, where OA = -2i - 4j. After 3 seconds, it is at B, where AB= 10i -40 j. Show that this is consistent with constant acceleration a. Find a and u
I need help on this aswell
To show that the motion is consistent with constant acceleration, we need to check if the acceleration is constant throughout the motion.
The position vector at any time t is given by the equation:
r(t) = r0 + ut + (1/2)at^2
Where:
- r0 is the initial position vector (at t = 0)
- u is the initial velocity vector (at t = 0)
- a is the constant acceleration vector
Given that at t = 0, the object is at the origin O, we have:
r0 = 0
Now, let's consider the position vector at t = 2 seconds, A:
r(2) = 0 + u(2) + (1/2)a(2)^2
r(2) = 2u + 2a
Given that OA = -2i - 4j, we have:
r(2) = -2i - 4j
Comparing the components, we get:
2u + 2a = -2i --(1)
2u + 2a = -4j --(2)
Now, let's consider the position vector at t = 3 seconds, B:
r(3) = 0 + u(3) + (1/2)a(3)^2
r(3) = 3u + (9/2)a
Given that AB = 10i - 40j, we have:
r(3) = 10i - 40j
Comparing the components, we get:
3u + (9/2)a = 10i --(3)
3u + (9/2)a = -40j --(4)
Now, we have a system of equations (1), (2), (3), and (4) that we can solve to find the acceleration vector a and initial velocity vector u.
Subtracting equation (4) from equation (2), we eliminate the variable u:
(-2i) - (-40j) = 2a
-2i + 40j = 2a
-2(i - 20j) = 2a
a = -i + 20j
Substituting this value of a into equation (1), we get:
2u + 2(-i + 20j) = -2i
2u - 2i + 40j = -2i
2u = 0
u = 0i
Therefore, the acceleration vector a is -i + 20j and the initial velocity vector u is 0i.
This shows that the motion is consistent with constant acceleration, and we have found the values of a and u.