The position of a particle moving along the x axis varies in time according to the expression x=4t^2+2 where x is in meters and t is in seconds. Evaluate it's position at the following times
a) Find the displacement from 0s to 3s
b) Find the average velocity from 0s to 3s
c) Find the instantaneous velocity at 0s and at 3s
d)Find the average acceleration from 0s to 3s
e)Find the instantaneous acceleration at 0s and at 3s
a. X1 = 4*0^2 + 2 = 2m. @ t = 0s.
X2 = 4*3^2 + 2 = 38m. @ 3s.
Disp. = X2-X1 = 38 - 2 = 36m.
b. V = 36m/3s. = 12m/s.
c. V(0s) = 0 m/s.
d. a = (12-0)/(3-0) = 4m/s^2.
e. a(0s) = 0m/s^2.
a(3s) = 4m/.s^2.
To find the answers to the given questions, let's start by understanding the given expression for the position of the particle:
x = 4t^2 + 2
a) The displacement from 0s to 3s can be found by calculating the change in position over that time interval. To do this, we substitute t = 3s into the expression:
x(3s) = 4(3^2) + 2
= 4(9) + 2
= 36 + 2
= 38 meters
So, the displacement from 0s to 3s is 38 meters.
b) The average velocity from 0s to 3s can be found by dividing the displacement by the time interval. In this case, the time interval is 3s:
Average Velocity = (Displacement)/(Time)
= (38 meters)/(3s)
= 12.67 m/s
Therefore, the average velocity from 0s to 3s is 12.67 m/s.
c) The instantaneous velocity at any given time can be found by taking the derivative of the position equation with respect to time. In this case, we can find the instantaneous velocity at t = 0s and t = 3s.
At t = 0s:
Velocity(0s) = d(x)/dt
= d(4t^2 + 2)/dt
= 8t
Plugging in t = 0s:
Velocity(0s) = 8(0)
= 0 m/s
So, the instantaneous velocity at 0s is 0 m/s.
At t = 3s:
Velocity(3s) = d(x)/dt
= d(4t^2 + 2)/dt
= 8t
Plugging in t = 3s:
Velocity(3s) = 8(3)
= 24 m/s
Therefore, the instantaneous velocity at 3s is 24 m/s.
d) The average acceleration from 0s to 3s can be found by dividing the change in velocity by the time interval. Since the velocity function is the derivative of the position function, the average acceleration can be calculated as:
Average Acceleration = (Change in Velocity)/(Time)
= (Velocity(3s) - Velocity(0s))/(3s - 0s)
Plugging in the previously calculated values:
Average Acceleration = (24 m/s - 0 m/s)/(3s - 0s)
= 24 m/s / 3s
= 8 m/s^2
Hence, the average acceleration from 0s to 3s is 8 m/s^2.
e) The instantaneous acceleration at any given time can be found by taking the derivative of the velocity equation with respect to time. Since the velocity equation is 8t, the derivative of velocity with respect to time will be the same:
Instantaneous Acceleration(0s) = d(Velocity)/dt
= d(8t)/dt
= 8
So, the instantaneous acceleration at 0s is 8 m/s^2.
Instantaneous Acceleration(3s) = d(Velocity)/dt
= d(8t)/dt
= 8
Hence, the instantaneous acceleration at 3s is also 8 m/s^2.
To summarize:
a) The displacement from 0s to 3s is 38 meters.
b) The average velocity from 0s to 3s is 12.67 m/s.
c) The instantaneous velocity at 0s is 0 m/s, and at 3s is 24 m/s.
d) The average acceleration from 0s to 3s is 8 m/s^2.
e) The instantaneous acceleration at 0s is 8 m/s^2, and at 3s is also 8 m/s^2.