# Solved Problems

## 2. Vector kinematics

### Problem 3

The velocity of a particle moving on the plane is given by the expression: (SI units). At the initial moment the particle is on the axis with position .

1. Find the moment when it passes through the axis and its distance from the origin at that moment.
2. Compute the acceleration at and at the moment when it passes through the axis.

The expression of the position is obtained by adding the initial position plus the integral of the velocity, with respect to time, from the initial time 0 to any time

(%i1) v: [3*exp(-2*t), -5*exp(-t)]\$
(%i2) r0: [0, 2]\$
(%i3) r: r0 + integrate(v,t,0,t);
(%o3)
(%i4) float(solve(r[2]=0, t));
(%o4)
(%i5) float(subst(%,r));
(%o5)

That is, the particle passes through the at , when its distance from the origin is 0.96 m.

(%i6) a: diff(v,t);
(%o6)
(%i7) subst(t=0,a);
(%o7)
(%i8) subst(%o4,a);
(%o8)

The acceleration vector at the initial time is  m/s2 and when it passes through the axis it is  m/s2.

### Problem 6

A pebble rolls down the roof of a house, which is inclined 20° from the horizontal. At the moment when the pebble leaves the roof and starts falling freely, it has a speed of 4 m/s and it is 6 m above the ground. Assuming that the air resistance is negligible,

1. Find the time it takes to hit the ground, from the moment when it leaves the roof.
2. What is the horizontal displacement of the pebble during its free fall?
3. Compute the angle between the vertical and the velocity of the pebble at the moment when it hits the ground.

With the axis in the horizontal and the axis in the vertical, the initial velocity vector is (SI units)

The starting position is and the acceleration vector is constant: . The expression for the velocity vector is obtained by integrating the acceleration vector from = 0 to any other time

And the expression of the position vector is obtained by integrating this expression for the velocity from = 0 to any other time

The time it takes to hit the ground is the time for which the component of the position is zero

And the horizontal distance is the value of the component of the position at that time

The velocity vector at the instant it hits the ground is

And the angle that the velocity makes with the vertical is the inverse tangent of the component divided by the absolute value of the component

### Problem 7

A boat transports passengers across a river, following the shortest path of 1.5 km between the two river banks. When the boat's motor works at its full power, the trip takes 20 minutes on a day when the speed of the river current is 1.2 m/s. Find the speed of the boat that day: (a) relative to the ground and (b) relative to the water. (c) Find the minimum time that it would take the same trip on a day when the speed of the river current was 0.8 m/s.

The boat crosses the river in a straight line, perpendicular to the banks and with 1500 m of length, with constant speed; therefore, the speed of the boat is:

The velocity vector of the boat relative to the water, , plus the velocity vector of the water, is equal to the velocity vector of the boat. Since the boat crosses the river in the direction perpendicular to the banks, the velocity of the water is perpendicular to the velocity of the boat, and these two velocities are the sides in a right triangle where is the hypotenuse, as shown in the figure.

Therefore, the speed of the boat relative to the water is m/s. The maximum speed of the boat relative to the water depends only on the operation of the engine and therefore will be the same on the day the current has speed 0.8 m/s, so the maximum speed of the boat on that day will be m/s. At that speed, the minimum time required to cross the river is 1500/1.537 = 975.9 s, namely, 16 minutes and 16 seconds.

### Problem 10

Three cylinders A, B and C are hanged by the two-pulley system shown in the figure. At a given instant, the velocity of block A is  m/s, upwards, and its acceleration is m/s2, downwards; at the same instant, the velocity and acceleration of block C are:  m/s, downwards, m/s2, upwards. Find the velocity and acceleration of block B, at the same instant, and say whether they are in the upwards or downwards direction.

Let us define 4 variables , , and to measure the positions of the cylinders and the moving pulley, relative to something fixed, for example the ceiling, as shown in the figure on the side.

Since cylinder A and the moving pulley are connected by a string with constant length,

And the link of cylinders B and C with another string of constant length, passing through the moving pulley, implies:

Differentiating these two equations with respect to time, we obtain the relations among the velocities:

As the distances increase when the objects move down, then the downward velocities are positive and upward ones are negative. Thus, the velocities given in the statement of the problem are and , and the above equation leads to ; namely, cylinder B moves downwards with speed of 5 m/s.

Differentiating again the relation among the velocities, we obtain the relation among the accelerations:

And replacing the given values, and , it is found that ; therefore, the acceleration of cylinder B is zero.

### Problem 11

For the system shown in the figure, find the relation between the velocities and accelerations of the bar A and the cylinder B, assuming that the bar A only moves up or down, remaining always in a horizontal position.

There are two distinct motions: the vertical motion of the bar, together with the two moving pulleys, and the vertical motion of the cylinder. These two motions are related to the change of the vertical positions of the bar and the cylinder, relative to some fixed reference object. Using as reference the ceiling let the positions of the bar and the cylinder be and , as shown in the following figure.

The vertical distance between either one the moving pulleys and one of the fixed pulleys is minus a constant. Thus, the total length of the string linking the cylinder to the ceiling is

Differentiating this equation with respect to time we obtain

And differentiating again