# Solved Problems

## 1. Kinematics

### Problem 2

The acceleration of object that moves along the axis is = −4 m/s2. If at = 0, the velocity is = +24 m/s and the object is at the position = 0, find its velocity and position at = 8 s and the total distance traveled by the object from = 0 to = 8 s.

To calculate the velocity at = 8, the constant value of the acceleration is substituted in the equation that relates the acceleration to the velocity and the time

Separating variables and integrating, we find the speed at = 8

To calculate the final position, we replace the acceleration value in the equation that relates the acceleration to the velocity and position

Separating variables and integrating,

The negative value of the final velocity means that the object has moved to a point where it stopped and at = 8 it's moving back in the direction of the origin. To calculate the total distance traveled one must determine the position of the point where it stopped

Thus, the object moved from = 0 to = 72 and it then traveled another 8 meters to = 64. The total distance traveled was then 80 m.

### Problem 3

At = 0 an object is at rest in the position = 5 cm in a path. From that moment, the object starts to move in the positive direction of the path, stopping again at . The expression for the tangential acceleration between and is , where time is measured in seconds and the acceleration in cm/s2. Find: (a) The instant when the object stops. (b) The position along the path at that instant.

The expression given for the tangential acceleration allows us to solve the equation

We then separate variables, integrate from 0 to and integrate from zero to zero again; the following Maxima commands can be used:

(%i1) integrate(9-3*t^2, t, 0, t1) = integrate(1, v, 0, 0);
(%o1)
(%i2) solve(%);
(%o2)

The result is then = 3 s. To find the position as a function of time, it is necessary to know the expression for the velocity as a function of time, which can be obtained by repeating the same integrals in (%i1), but leaving the upper limits as variables and .

(%i3) integrate(9-3*t^2, t, 0, t) = integrate(1, v, 0, v);
(%o3)

That will allow us to solve the equation

Separating variables, integrating from = 0 to = 3 and from the initial position = 5 to the final position , we get

(%i4) integrate (lhs(%), t, 0, 3) = integrate(1, s, 5, s1);
(%o4)
(%i5) solve(%);
(%o5)

The final position is = 25.25 cm.

### Problem 5

The acceleration of an object that oscillates on the axis is , where is a positive constant. Find:

1. The value that must have so that = 15 m/s at = 0 and = 0 at = 3 m.
2. The velocity of the object at = 2 m.

The expression for the projection of the acceleration as a function of the coordinate allows us to solve the following equation

Separating variables and integrating between the two values ​​of and given, we get the value of the constant (SI units):

To find the velocity in = 2, the same integrals are solved, but now the value of is known and the unknown variable is the velocity in = 2

The positive and negative values ​​of are due to the fact that, as the particle oscillates, it passes through = 2, sometimes in the positive direction and sometimes in the negative direction.

### Problem 7

The square of the velocity of an object decreases linearly as a function of its position on the trajectory, as shown in the figure. Compute the distance traveled by the object during the last two seconds before it reaches point B.

The equation of the straight line in the plot is:

(%i6) eq1: v^2 = m*s + b;
(%o6)

The two parameters and can be found using the coordinates of points A and B in the plot

(%i7) solve ([subst([s=100,v^2=2500],eq1), subst([s=400,v^2=900],eq1)]);
(%o7)
(%i8) subst (%[1], eq1);
(%o8)
(%i9) solve (%, v);
(%o9)

Since it is said in the statement of the problem that the square of the velocity is decreasing, it 's concluded that the object passes first through a nd then through B. Thus, the object is moving in the positive direction and the velocity is positive, which is the second solution in the list above.

(%i10) eq2: %[2];
(%o10)

This last expression must be equal to , and separation of variables leads to:

Assuming that = 0 in the point C where the object is 2 seconds before it reaches point B, then will be equal to 2 at point B, so the limits for the time integral will be from 0 to 2 and the limits for the integral will be and 400. We then have,

(%i11) integrate (1, t, 0, 2) = integrate (1/rhs(eq2), s, sC, 400);
Is sC - 400 positive, negative or zero?
negative;
(%o11)

The negative option was chosen, because we now that is smaller than , which is 400, but in fact whatever we answered in this case would have led to the same result. The position of the point C, 2 seconds before arriving at B, is then:

(%o12) solve (%);
(%o12)

The total distance traveled during the last two seconds before the object arrives at B is then , which in meters is equal to,

(%i13) float (400 - rhs(%[1]));
(%o13)    65.33

### Problem 11

A marble is thrown over the horizontal surface on the top of some stairs, and it leaves the top of the stairs with horizontal speed of 3 m/s. Each step goes down 18 cm and is 30 cm wide. On which step will the marble land?

The velocity of the projection of the marble in the horizontal axis remains constant and equal to its initial value, = 3 (SI units). The horizontal position of the marble at any time is then , taking = 0, when it leaves the top of the stairs.

The velocity of the projection on the vertical axis, at = 0, is = 0, because the marble leaves the top of the stairs in the horizontal direction. The acceleration of the motion of the vertical projection is constant and equal to the acceleration of gravity: = -9.8. Integrating with respect to , from 0 to , we get:

And integrating this expression with respect to , from 0 to , we get the expression for the height , measured from the top of the stairs:

Eliminating between this equation and the expression , we get the equation of the parabolic trajectory of the marble, with origin at the edge of the top of the stairs:

From that same origin, the edges of the successive steps are at (0.3, −0.18), (0.6, −0.36), (0.9, −0.48), … Therefore, if we compute the sequence , for = 1, 2, 3,…, we'll have the height of the marble above each step, as it passes through the edge ot each step. The value of for which that height becomes negative, will tell us that the marble didn't clear that step but it rather landed on it. The first 10 elements of the sequence can be computed in Maxima with the following command:

(%i14) makelist (-(4.9/9)*(0.3*n)^2 + 0.18*n, n, 1, 10);
(%o14) [0.131, 0.164, 0.099, −0.064, −0.325, −0.684, −1.141, −1.696, −2.349, −3.1]

Since the sequence becomes negative at the fourth element, we conclude that the marble hits the fourth step.