Displacement of the Object

To find the displacement covered by the object at the end of the third second of motion, we need to integrate the velocity function with respect to time.

The given velocity function is:

v(t) = 5t - 2t^2 + 4

To find the displacement, we integrate the velocity function over the time interval from 0 to 3 seconds:

∫[0 to 3] (5t - 2t^2 + 4) dt

Let's integrate each term separately:

∫[0 to 3] 5t dt = (5/2)t^2 | [0 to 3] = (5/2)(3)^2 - (5/2)(0)^2 = 22.5 m

∫[0 to 3] -2t^2 dt = (-2/3)t^3 | [0 to 3] = (-2/3)(3)^3 - (-2/3)(0)^3 = -18 m

∫[0 to 3] 4 dt = 4t | [0 to 3] = 4(3) - 4(0) = 12 m

Adding up the integrals, we get:

Displacement = 22.5 - 18 + 12 = 16.5 m

Therefore, the object covers a displacement of 16.5 meters at the end of the third second of motion.

Acceleration of the Object

To find the acceleration as a function of time, we differentiate the velocity function with respect to time.

The given velocity function is:

v(t) = 5t - 2t^2 + 4

Differentiating each term separately:

d/dt (5t) = 5

d/dt (-2t^2) = -4t

d/dt (4) = 0

Adding up the derivatives, we get:

Acceleration = 5 - 4t

Interpreting the Acceleration

The acceleration function we obtained is a linear function of time, given by:

a(t) = 5 - 4t

This means that the acceleration is changing at a constant rate of -4 m/s^2 per second. The negative sign indicates that the acceleration is decreasing over time.

Since the acceleration is negative, it implies that the object is slowing down. As time progresses, the rate of deceleration increases. This can be seen from the negative coefficient (-4) multiplying the time variable (t) in the acceleration function.

In summary, the object has a displacement of 16.5 meters at the end of the third second of motion. The acceleration of the object is given by the function a(t) = 5 - 4t, indicating that the object is slowing down at a constant rate.