a particle moving in the xy plane has velocity components dx/dt =2t and dy/dt = 4. If the particle starts from the origin at t = 0,magnitude of its displacement at t=2 is
To find the displacement of the particle, we need to integrate the velocity components with respect to time.
Given: dx/dt = 2t and dy/dt = 4
Integrating dx/dt with respect to t, we get:
∫dx = ∫2t dt
x = t^2 + C1
Integrating dy/dt with respect to t, we get:
∫dy = ∫4 dt
y = 4t + C2
Since the particle starts from the origin at t = 0, we have x(0) = 0 and y(0) = 0.
Plugging in these values into the equations, we get:
0 = 0^2 + C1
C1 = 0
0 = 4(0) + C2
C2 = 0
Therefore, the equations for x and y become:
x = t^2
y = 4t
To find the magnitude of the displacement at t = 2, we can find the distance between the point (x,y) and the origin (0,0).
Using the Pythagorean theorem, the magnitude of the displacement is:
d = sqrt((x-0)^2 + (y-0)^2)
d = sqrt(x^2 + y^2)
Plugging in the equations for x and y, we get:
d = sqrt((t^2)^2 + (4t)^2)
d = sqrt(t^4 + 16t^2)
Plugging in t = 2, we get:
d = sqrt((2^4) + 16(2^2))
d = sqrt(16 + 64)
d = sqrt(80)
Therefore, the magnitude of the displacement at t = 2 is sqrt(80).
To find the magnitude of the displacement at t = 2, we need to integrate the velocity components with respect to time to get the position components.
Given:
dx/dt = 2t
dy/dt = 4
Integrating both equations, we get:
x = ∫(2t) dt = t^2 + C1 --------- (Equation 1)
y = ∫4 dt = 4t + C2 --------- (Equation 2)
Now, let's find the constant values C1 and C2 using the initial conditions. Since the particle starts from the origin (x = 0, y = 0) at t = 0, we can substitute these values into the equations:
x = 0 = (0^2) + C1
C1 = 0
y = 0 = 4(0) + C2
C2 = 0
So, the position components become:
x = t^2 --------- (Equation 3)
y = 4t --------- (Equation 4)
Now, to find the magnitude of displacement at t = 2, we substitute t = 2 into equations (3) and (4):
x = (2)^2 = 4
y = 4(2) = 8
Using the Pythagorean theorem, the magnitude of displacement is given by:
Magnitude of displacement = √(x^2 + y^2)
= √(4^2 + 8^2)
= √(16 + 64)
= √80
= 4√5
Therefore, the magnitude of the displacement at t = 2 is 4√5.
dx/dt=2t, integrate, x=t^2+Cx
dy/dt=4, integrate, y=4t+Cy
We know x(0)=y(0)=0 => Cx=Cy=0
The trajectory of the particle is therefore
<t^2, 4t>
Guess you can find displacement at t=2 with the above information.