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.

intigrate