At t= 0, a particle starts from rest at x= 0, y= 0, and moves in the xy plane with an acceleration ->a (vector) = (4.0ihat+ 3.0jhat)m/s^2. Assume t is in seconds.
Determine the x component of velocity as a function of time t.
Determine the y component of velocity as a function of time t.
Determine the speed of the particle as a function of time t.
Determine the position of the particle as a function of time t.
To determine the x component of velocity as a function of time t, we need to integrate the x component of acceleration with respect to time.
Given ->a = (4.0î + 3.0ĵ) m/s², the x component of acceleration is 4.0 m/s².
To integrate the x component of acceleration with respect to time, we can use the formula for the x component of velocity:
v_x(t) = ∫(a_x dt)
Since the x component of acceleration, a_x, is constant, the integral becomes:
v_x(t) = a_x * t
Therefore, the x component of velocity as a function of time t is given by:
v_x(t) = 4.0t
To determine the y component of velocity as a function of time t, we can follow the same process as above. The y component of acceleration is 3.0 m/s².
Therefore, the y component of velocity as a function of time t is given by:
v_y(t) = 3.0t
To determine the speed of the particle as a function of time t, we can use the Pythagorean theorem. The speed of the particle is equal to the magnitude of the velocity vector.
The velocity vector is given by:
v(t) = (v_x(t), v_y(t)) = (4.0t, 3.0t)
The magnitude of the velocity vector can be calculated using the formula:
speed = |v(t)| = sqrt(v_x(t)² + v_y(t)²)
Therefore, the speed of the particle as a function of time t is given by:
speed = sqrt((4.0t)² + (3.0t)²) = sqrt(16t² + 9t²) = sqrt(25t²) = 5t
Finally, to determine the position of the particle as a function of time t, we can integrate the x and y components of velocity.
Given the initial conditions x(0) = 0 and y(0) = 0, we can find the position functions by integrating the corresponding velocity components.
x(t) = ∫(v_x(t) dt) = ∫(4.0t dt) = 2.0t² + C1
y(t) = ∫(v_y(t) dt) = ∫(3.0t dt) = 1.5t² + C2
The constants of integration C1 and C2 represent the initial position of the particle and can be found using the initial conditions x(0) = 0 and y(0) = 0.
Therefore, the position of the particle as a function of time t is given by:
x(t) = 2.0t²
y(t) = 1.5t²
Note: The units for velocity are m/s, and the units for position are meters.
How would you integrate them?
4.0 is the acceleration in the x direction (with the i-hat unit vector)) and 3.0 is the acceleration in the y direction (with the j-hat unit vector).
Vx = 4.0 t Vy = 3.0 t
V = sqrt(V2^2 + Vy^2) = 5 t
Integrate Vx and Vy for the x and y locations vs. t.