A particle is moving in x-y plane such that its velocity is given by vx = 3t2 and vy= 2t .fine average velocity of particle at t=2 seconds
To find the average velocity of the particle at t = 2 seconds, we first need to find the displacement of the particle during the time interval from t = 0 seconds to t = 2 seconds.
The displacement in the x-direction (Δx) can be found by integrating the x-component of the velocity function (vx) with respect to time:
Δx = ∫(vx)dt = ∫(3t^2)dt
Integrating 3t^2 with respect to t, we get:
Δx = t^3 + C
Next, we need to find the value of the constant C. Since we want to find the displacement from t = 0 seconds to t = 2 seconds, we can substitute t = 0 seconds into the equation:
0 = (0)^3 + C
C = 0
Therefore, the displacement in the x-direction is:
Δx = t^3
Similarly, we can find the displacement in the y-direction (Δy) by integrating the y-component of the velocity function (vy) with respect to time:
Δy = ∫(vy)dt = ∫(2t)dt
Integrating 2t with respect to t, we get:
Δy = t^2 + D
To find the value of the constant D, we can again substitute t = 0 seconds into the equation:
0 = (0)^2 + D
D = 0
Therefore, the displacement in the y-direction is:
Δy = t^2
Now, we can calculate the displacement vector (Δr) using the Pythagorean theorem:
Δr = √(Δx^2 + Δy^2)
= √((t^3)^2 + (t^2)^2)
= √(t^6 + t^4)
Finally, we can calculate the average velocity (Vavg) by dividing the displacement vector by the time interval:
Vavg = Δr / Δt
= (√(t^6 + t^4)) / (2 - 0)
= (√(t^6 + t^4)) / 2
Substituting t = 2 seconds, we get:
Vavg = (√((2^6) + (2^4))) / 2
= (√(64 + 16)) / 2
= (√80) / 2
= (√16 * 5) / 2
= (4√5) / 2
= 2√5
Therefore, the average velocity of the particle at t = 2 seconds is 2√5, where √ represents the square root.