Suppose the position of an object is given by ->r(vector) = (3.0t^2*ihat - 6.0t^3*jhat)m. Where t in seconds.
Determine its velocity ->v as a function of time t.
Determine its acceleration ->a as a function of time t.
Determine ->r at time t = 2.5 s.
Determine ->v at time t = 2.5s.
I don't really understand the vector units and the answers require the i^ and j^ vector units. Please explain the significance, my prof didn't do justice in class.
R = 3 t^2 i -6 t^3 j
V = dR/dt = 6 t i - 18 t^2 j
A = d^2R/dt^2 = dV/dt = 6 i - 36 t j
use 2.5 for t to get numbers
pweasey-weasy?
thank you damon
In order to understand the vector units and the significance of the i^ and j^ vector units, we need to understand the concept of vectors in physics.
In physics, a vector is a quantity that has both magnitude and direction. It is commonly represented as an arrow, where the length of the arrow represents the magnitude of the vector and the direction of the arrow represents the direction of the vector.
In this case, the vector r represents the position of the object. It is given by r = (3.0t^2)i^ - (6.0t^3)j^, where i^ and j^ are the unit vectors in the x and y directions respectively.
The unit vector i^ represents the positive x direction, while the unit vector j^ represents the positive y direction. So, when we have a term multiplied by i^, it represents the magnitude of the vector in the x direction, and when we have a term multiplied by j^, it represents the magnitude of the vector in the y direction.
Now, let's determine the velocity v as a function of time t. Velocity is defined as the rate of change of position with respect to time, so we can find it by taking the derivative of the position vector r with respect to time.
v = d(r)/dt = (d(3.0t^2)/dt)i^ - (d(6.0t^3)/dt)j^
Taking the derivatives, we get:
v = (6.0t)i^ - (18.0t^2)j^
This is the velocity vector as a function of time.
Next, let's determine the acceleration a as a function of time t. Acceleration is defined as the rate of change of velocity with respect to time, so we can find it by taking the derivative of the velocity vector v with respect to time.
a = d(v)/dt = (d(6.0t)/dt)i^ - (d(18.0t^2)/dt)j^
Taking the derivatives, we get:
a = 6.0i^ - (36.0t)j^
This is the acceleration vector as a function of time.
Now, let's determine r at time t = 2.5 s. Substitute t = 2.5 s into the position vector r:
r(2.5) = (3.0(2.5)^2)i^ - (6.0(2.5)^3)j^
Calculating the values, we get:
r(2.5) = (18.75)i^ - (93.75)j^
Therefore, the position vector r at time t = 2.5 s is (18.75)i^ - (93.75)j^.
Finally, let's determine v at time t = 2.5 s. Substitute t = 2.5 s into the velocity vector v:
v(2.5) = (6.0(2.5))i^ - (18.0(2.5)^2)j^
Calculating the values, we get:
v(2.5) = (15.0)i^ - (112.5)j^
Therefore, the velocity vector v at time t = 2.5 s is (15.0)i^ - (112.5)j^.