Explain the solution of problems related to situations that involve the use of differential and integral calculus. Instructions 1. The following equation represents the speed v measured in m/s with respect to the time t measured in seconds (s) of a ball moving on a horizontal surface. v=-0.12t+1.64, From this, find: a) The position function b) The expression for the acceleration of the ball.

To find the position function, we need to integrate the given speed function with respect to time.

a) The position function can be found by integrating the speed function, which is given by:

$$v = -0.12t + 1.64$$

To integrate this function, we treat v as a function of t and integrate each term separately:

$$\int v dt = \int (-0.12t + 1.64) dt$$

Integrating the first term:

$$\int (-0.12t) dt = -0.12 \int t dt$$

Using the power rule of integration, the integral of t with respect to t is:

$$\int t dt = \frac{t^2}{2}$$

So, the integral of -0.12t with respect to t is:

$$-0.12 \int t dt = -0.12 \cdot \frac{t^2}{2} = -0.06t^2$$

Integrating the second term:

$$\int 1.64 dt = 1.64t$$

Putting it all together, we get the position function:

$$x(t) = -0.06t^2 + 1.64t + C$$

where C is the constant of integration.

b) The expression for the acceleration of the ball can be found by differentiating the speed function with respect to time.

Given the speed function:

$$v = -0.12t + 1.64$$

We can find the acceleration function by taking the derivative of v with respect to t:

$$a = \frac{dv}{dt}$$

Differentiating each term of the speed function:

$$\frac{d}{dt} (-0.12t) = -0.12$$

$$\frac{d}{dt} (1.64) = 0$$

Summing up the derivatives, we get the acceleration function:

$$a(t) = -0.12$$

Answer:
a) The position function is given by: $$x(t) = -0.06t^2 + 1.64t + C$$
b) The expression for the acceleration of the ball is: $$a(t) = -0.12$$