In a particular Cartesian coordinate system, the y and z-components of the velocity of a point-like object are zero and the x-component varies as given by the following function:
vx(t) = 2 + 5t,
where t is in seconds and vx is in meters per second.
If the particle was at x = 0 at t = 0, ?nd the position, in meters, of the particle at t = 1.2 s.
x(t_1)= ?
If the particle was at x = 0 at t = 0, ?nd the position, in meters, of the particle at t = 2.2 s.
x(t_2) = ?
Well, well, well, let's calculate the positions of this point-like object using some good old mathematics!
For the first part, to find the position at t = 1.2s, we can integrate the velocity function with respect to time. So, integrating vx(t) = 2 + 5t gives us x(t) = 2t + (5/2)t^2 + C. Now, since the particle was at x = 0 at t = 0, we can substitute these values into the equation to find the constant, C.
0 = 2(0) + (5/2)(0)^2 + C
0 = 0 + 0 + C
C = 0
Substituting C = 0 back into the equation, we get x(t) = 2t + (5/2)t^2, and we can plug in t = 1.2s to find the position:
x(1.2) = 2(1.2) + (5/2)(1.2)^2
x(1.2) = 2.4 + (5/2)(1.44)
x(1.2) = 2.4 + (5/2)(1.44)
x(1.2) = 2.4 + 3.6
x(1.2) = 6 meters
So, the position of the particle at t = 1.2s is 6 meters.
Now, for the second part, we can use the same equation x(t) = 2t + (5/2)t^2. Substituting t = 2.2s, we can find the position:
x(2.2) = 2(2.2) + (5/2)(2.2)^2
x(2.2) = 4.4 + (5/2)(4.84)
x(2.2) = 4.4 + 6.1
x(2.2) = 10.5 meters
Therefore, the position of the particle at t = 2.2s is 10.5 meters.
Hope that puts a smile on your face!
To find the position of the particle at different times, we can integrate the velocity function with respect to time.
First, we need to integrate the given velocity function vx(t) = 2 + 5t with respect to t to obtain the position function x(t).
∫(2 + 5t) dt = 2t + (5/2)t^2 + C
Since the particle was at x = 0 at t = 0, we can solve for the constant C by substituting these values into the position function:
x(0) = 2(0) + (5/2)(0)^2 + C
x(0) = C
So, C = 0.
The position function is:
x(t) = 2t + (5/2)t^2
Now we can find the position of the particle at specific times.
1. At t = 1.2 s:
x(t_1) = 2(1.2) + (5/2)(1.2)^2
x(t_1) = 2.4 + (5/2)(1.44)
x(t_1) = 2.4 + 3.6
x(t_1) = 6 meters
The position of the particle at t = 1.2 s is 6 meters.
2. At t = 2.2 s:
x(t_2) = 2(2.2) + (5/2)(2.2)^2
x(t_2) = 4.4 + (5/2)(4.84)
x(t_2) = 4.4 + 12.1
x(t_2) = 16.5 meters
The position of the particle at t = 2.2 s is 16.5 meters.
To find the position of the particle at a given time, we can integrate the x-component of velocity with respect to time.
Given that the x-component of velocity is given by vx(t) = 2 + 5t, we can integrate this function to find the x-position as a function of time.
To find the position at t = 1.2 seconds, we can substitute t = 1.2 into the integrated equation.
x(t) = ∫(2 + 5t) dt
= 2t + (5/2)*t^2 + C
Since the particle was at x = 0 at t = 0, we can solve for the constant C.
0 = 2(0) + (5/2)(0)^2 + C
= 0 + 0 + C
= C
Therefore, C = 0.
Now we can substitute t = 1.2 into the equation to find the x-position.
x(1.2) = 2(1.2) + (5/2)*(1.2)^2
= 2.4 + (5/2)*1.44
= 2.4 + 3.6
= 6 meters
So the position of the particle at t = 1.2 seconds is x(t₁) = 6 meters.
Similarly, to find the position at t = 2.2 seconds, we can substitute t = 2.2 into the integrated equation.
x(t) = 2t + (5/2)*t^2 + C
Since the particle was at x = 0 at t = 0, we can solve for the constant C.
0 = 2(0) + (5/2)(0)^2 + C
= 0 + 0 + C
= C
Therefore, C = 0.
Now we can substitute t = 2.2 into the equation to find the x-position.
x(2.2) = 2(2.2) + (5/2)*(2.2)^2
= 4.4 + (5/2)*4.84
= 4.4 + 12.1
= 16.5 meters
So the position of the particle at t = 2.2 seconds is x(t₂) = 16.5 meters.