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Solve Rectilinear Translation Problems in Physics with This PDF 225l of Solutions

Rectilinear Translation Problems with Solution PDF

Rectilinear translation is a type of motion along a straight line, which is often encountered in physics and engineering. Solving rectilinear translation problems involves applying the principles of kinematics, such as displacement, velocity, acceleration and free fall. In this article, I will explain what rectilinear translation is, how to solve some common problems using equations and graphs, and where to find some useful resources for more practice and learning.

What is rectilinear translation?

Rectilinear translation is the motion of an object along a straight line. For example, a car moving on a straight road, a stone thrown vertically upward or downward, or a bullet fired from a gun are all examples of rectilinear translation. In this type of motion, the direction of the object does not change, only its speed or magnitude of velocity may change.

Why is it important to study rectilinear translation?

Rectilinear translation is important to study because it is one of the simplest forms of motion that can be analyzed using basic concepts of kinematics. Kinematics is the branch of physics that deals with the description of motion without considering its causes or effects. By studying rectilinear translation, we can learn how to describe the position, velocity and acceleration of an object at any given time using mathematical equations or graphical representations. These skills are essential for solving more complex problems involving forces, energy, momentum or rotation.

Basic concepts

Displacement

Displacement is the net change in position of an object during its motion. It is a vector quantity, which means it has both magnitude and direction. Displacement can be positive or negative depending on the direction of motion relative to a reference point. Displacement can be calculated by subtracting the initial position from the final position of an object:

\Delta r = r_2 - r_1

where \Delta r is the displacement vector, r_2 is the final position vector and r_1 is the initial position vector.

If we assume that the motion is along a horizontal axis (x-axis), then we can write:

\Delta x = x_2 - x_1

where \Delta x is the displacement along the x-axis, x_2 is the final x-coordinate and x_1 is the initial x-coordinate.

Note that displacement is not the same as total distance traveled. Total distance is the length of the path followed by the object, which may be longer than the displacement if the object changes direction or moves in a curved path.

Velocity

Velocity is the rate of change of position of an object. It is also a vector quantity, which means it has both magnitude and direction. Velocity can be positive or negative depending on the direction of motion relative to a reference point. Velocity can be calculated by dividing the displacement by the time interval during which the displacement occurs:

v = \frac\Delta r\Delta t

where v is the velocity vector, \Delta r is the displacement vector and \Delta t is the time interval.

If we assume that the motion is along a horizontal axis (x-axis), then we can write:

v_x = \frac\Delta x\Delta t

where v_x is the velocity along the x-axis, \Delta x is the displacement along the x-axis and \Delta t is the time interval.

There are two types of velocity: average velocity and instantaneous velocity. Average velocity is the ratio of the total displacement to the total time interval during which the motion occurs. Instantaneous velocity is the velocity of an object at a specific instant of time or at a specific point of its path. Instantaneous velocity can be obtained by taking the limit of average velocity as the time interval approaches zero:

v = \lim_\Delta t \to 0 \frac\Delta r\Delta t = \fracdrdt

The derivative of position with respect to time gives the instantaneous velocity.

Acceleration

Acceleration is the rate of change of velocity of an object. It is also a vector quantity, which means it has both magnitude and direction. Acceleration can be positive or negative depending on whether the velocity increases or decreases in magnitude or changes direction. Acceleration can be calculated by dividing the change in velocity by the time interval during which the change in velocity occurs:

a = \frac\Delta v\Delta t

where a is the acceleration vector, \Delta v is the change in velocity vector and \Delta t is the time interval.

If we assume that the motion is along a horizontal axis (x-axis), then we can write:

a_x = \frac\Delta v_x\Delta t

where a_x is the acceleration along the x-axis, \Delta v_x is the change in velocity along the x-axis and \Delta t is the time interval.

There are two types of acceleration: average acceleration and instantaneous acceleration. Average acceleration is the ratio of the total change in velocity to the total time interval during which the motion occurs. Instantaneous acceleration is the acceleration of an object at a specific instant of time or at a specific point of its path. Instantaneous acceleration can be obtained by taking the limit of average acceleration as the time interval approaches zero:

a = \lim_\Delta t \to 0 \frac\Delta v\Delta t = \fracdvdt

The derivative of velocity with respect to time gives the instantaneous acceleration.

Free fall

Free fall is a special case of rectilinear translation where an object falls under the influence of gravity only, without any other forces acting on it. In free fall, an object accelerates downward at a constant rate, which is equal to the acceleration due to gravity, denoted by g. The value of g varies slightly depending on location, but it can be approximated as 9.8 m/s or 32 ft/s.

In free fall, we can apply the following equations of motion for constant acceleration:

• v = v_0 + gt, where v is the final velocity, v_0 is the initial velocity and t is the time elapsed.

• y = y_0 + v_0t + \frac12gt^2, where y is the final position, Continuing the article: y_0 is the initial position and g is the acceleration due to gravity.

• v^2 = v_0^2 + 2g(y - y_0), where v is the final velocity, v_0 is the initial velocity, y is the final position and y_0 is the initial position.

Note that these equations are valid only for free fall motion, where a = g. If there are other forces acting on the object, such as air resistance or friction, then these equations do not apply.

Solving problems

Equations of motion

To solve rectilinear translation problems using equations of motion, we need to follow these steps:

• Identify the given and unknown quantities in the problem. Assign symbols and units to them.

• Choose a coordinate system and a reference point. Define the positive and negative directions.

• Select an equation of motion that relates the given and unknown quantities. If there are more than one unknowns, use more than one equation.

• Substitute the given values into the equation(s) and solve for the unknown(s). Check the units and signs of the answer(s).

• Verify that the answer(s) make sense physically and logically.

Graphical method

To solve rectilinear translation problems using graphical method, we need to follow these steps:

• Identify the given and unknown quantities in the problem. Assign symbols and units to them.

• Choose a coordinate system and a reference point. Define the positive and negative directions.

• Draw a position-time, velocity-time or acceleration-time graph that represents the motion of the object. Use a suitable scale and label the axes.

• Use the graph to find the unknown quantities by reading off values, calculating slopes or finding areas under curves.

• Verify that the answer(s) make sense physically and logically.

Example problems

Let's look at some example problems that illustrate how to use equations and graphs to solve rectilinear translation problems.

Example 1: A stone is thrown vertically upward with an initial velocity of 20 m/s. Find its maximum height, time to reach maximum height, total time in the air and velocity when it hits the ground.

Solution:

We can use equations of motion to solve this problem. Let's define up as positive and down as negative. The given and unknown quantities are:

• v_0 = +20 m/s

• v = ?

• y_0 = 0 m

• y = ?

• t = ?

• g = -9.8 m/s^2

To find the maximum height, we need to use the fact that at the highest point, the velocity of the stone is zero. So we can use the equation:

v^2 = v_0^2 + 2g(y - y_0)

Substituting v = 0, v_0 = +20 m/s, y_0 = 0 m and g = -9.8 m/s^2, we get:

(0)^2 = (20)^2 + 2(-9.8)(y - 0)

Solving for y, we get:

y = \frac(20)^22(9.8) \approx 20.4 m

The maximum height is about 20.4 m.

To find the time to reach maximum height, we can use another equation:

v = v_0 + gt

Substituting v = 0, v_0 = +20 m/s and g = -9.8 m/s^2, we get:

0 = 20 + (-9.8)t

Solving for t, we get:

t = \frac-20-9.8 \approx 2.04 s

The time to reach maximum height is about 2.04 s.

To find the total time in the air, we can use the fact that the stone takes the same time to go up as to come down, since the initial and final velocities are equal in magnitude but opposite in direction. So we can simply double the time to reach maximum height:

t_total = 2t \approx 4.08 s

The total time in the air is about 4.08 s.

To find the velocity when it hits the ground, we can use the same equation as before:

v^2 = v_0^2 + 2g(y - y_0)

Substituting y = 0 m, y_0 = 0 m, v_0 = +20 m/s and g = -9.8 m/s^2, we get:

v^2 = (20)^2 + 2(-9.8)(0 - 0)

Solving for v, we get:

v = \pm \sqrt(20)^2 = \pm 20 m/s

Since the stone is moving downward when it hits the ground, we choose the negative sign:

v = -20 m/s

The velocity when it hits the ground is -20 m/s.

Example 2: A ball is thrown upward with an initial velocity of 15 m/s from a height of 10 m above the ground. Draw a position-time graph and a velocity-time graph for the motion of the ball. Use the graphs to find the maximum height, time to reach maximum height, total time in the air and velocity when it hits the ground.

Solution:

We can use graphical method to solve this problem. Let's define up as positive and down as negative. The given and unknown quantities are:

• v_0 = +15 m/s

• v = ?

• y_0 = +10 m

• y = ?

• t = ?

• g = -9.8 m/s^2

Continuing the article: To draw a position-time graph, we need to plot the position of the ball at different times. We can use the equation:

y = y_0 + v_0t + \frac12gt^2

Substituting y_0 = +10 m, v_0 = +15 m/s and g = -9.8 m/s^2, we get:

y = 10 + 15t - 4.9t^2

This is a quadratic equation that gives a parabolic shape for the graph. The graph looks like this:

![position-time graph](https://i.imgur.com/6wZxqfD.png) The position-time graph shows that the ball starts at a height of 10 m, reaches a maximum height of about 17.6 m at about 1.5 s, and then falls back to the ground at about 3.1 s.

To draw a velocity-time graph, we need to plot the velocity of the ball at different times. We can use the equation:

v = v_0 + gt

Substituting v_0 = +15 m/s and g = -9.8 m/s^2, we get:

v = 15 - 9.8t

This is a linear equation that gives a straight line for the graph. The graph looks like this:

![velocity-time graph](https://i.imgur.com/7FyYsQd.png) The velocity-time graph shows that the ball starts with a positive velocity of 15 m/s, decreases linearly as it goes up, becomes zero at the maximum height, and then becomes negative as it falls down. The slope of the line is equal to the acceleration, which is constant and negative (-9.8 m/s).

To find the maximum height from the position-time graph, we need to find the vertex of the parabola. The vertex occurs when t = -\fracb2a, where b and a are the coefficients of t and t^2 in the quadratic equation. In this case, b = 15 and a = -4.9, so we get:

t = -\frac152(-4.9) \approx 1.53 s

This is the time to reach maximum height. To find the maximum height, we plug this value of t into the equation for y:

y = 10 + 15(1.53) - 4.9(1.53)^2 \approx 17.6 m

This is the maximum height.

To find the time to reach maximum height from the velocity-time graph, we need to find when the velocity becomes zero. This occurs when t = -\fracv_0g, where v_0 and g are the initial velocity and acceleration respectively. In this case, v_0 = 15 m/s and g = -9.8 m/s^2, so we get:

t = -\frac15-9.8 \approx 1.53 s

This is the same value as before.

To find the total time in the air from the position-time graph, we need to find when the ball hits the ground. This occurs when y = 0. So we set y = 0 in the equation for y and solve for t:

0 = 10 + 15t - 4.9t^2

This is a quadratic equation that can be solved by factoring, completing the square or using the quadratic formula. We will use the quadratic formula:

t = \frac-b \pm \sqrtb^2 - 4ac2a

where b = 15, a = -4.9 and c = 10. Substituting these values, we get:

t = \frac-15 \pm \sqrt15^2 - 4(-4.9)(10)2(-4.9)

Simplifying, we get:

t = \frac-15 \pm 19.6-9.8

There are two possible solutions: t \approx -0.36 s or t \approx 3.11 s. The negative solution does not make sense physically, since it implies the ball hit the ground before it was thrown. So we choose the positive solution:

t \approx 3.11 s

This is the total time in the air.

To find the velocity when it hits the ground from the velocity-time graph, we need to find the value of v when t = 3.11 s. We can use the equation for v and plug in this value of t:

v = 15 - 9.8(3.11) \approx -15.5 m/s

This is the velocity when it hits the ground.

Example 3: A bullet is shot vertically upward with a velocity of 300 m/s from a height of 2 m above the ground. Find its maximum height, time to reach maximum height, total time in the air and velocity when it hits the ground.

Solution:

We can use equations of motion to solve this problem. Let's define up as positive and down as negative. The given and unknown quantities are:

• v_0 = +300 m/s

• v = ?

• y_0 = +2 m

• y = ?

• t = ?

• g = -9.8 m/s^2

Continuing the article: To find the maximum height, we need to use the fact that at the highest point, the velocity of the bullet is zero. So we can use the equation:

v^2 = v_0^2 + 2g(y - y_0)

Substituting v = 0, v_0 = +300 m/s, y_0 = +2 m and g = -9.8 m/s^2, we get:

(0)^2 = (300)^2 + 2(-9.8)(y - 2)

Solving for y, we get:

y = \frac(300)^22(9.8) + 2 \approx 4614.3 m

The maximum height is about 4614.3 m.

To find the time to reach maximum height, we can use another equation:

v = v_0 + gt

Substituting v = 0, v_0 = +300 m/s and g = -9.8 m/s^2, we get:

0 = 300 + (-9.8)t

Solving for t, we get:

t = \frac-300-9.8 \approx 30.6 s

The time to reach maximum height is about 30.6 s.

To find the total time in the air, we can use the fact that the bullet takes the same time to go up as to come down, since the initial and final velocities are equal in magnitude but opposite in direction. So we can simply double the time to reach maximum height:

t_total = 2t \approx 61.2 s

The total time in the air is about 61.2 s.

To find the velocity when it hits the ground, we can use the same equation as before:

v^2 = v_0^2 + 2g(y - y_0)

Substituting y = 0 m, y_0 = +2 m, v_0 = +300 m/s and g = -9.8 m/s^2, we get:

v^2 = (300)^2 + 2(-9.8)(0 - 2)

Solving for v, we get:

v = \pm \sqrt(300)^2 + 39.2 \approx \pm 300.1 m/s

Since the bullet is moving downward when it hits the ground, we choose the negative sign:

v = -300.1 m/s

The velocity when it hits the ground is -300.1 m/s.

Conclusion

In this article, we have learned what rectilinear translation is and how to solve some common problems using equations and graphs. We have also seen some examples of rectilinear translation problems involving free fall motion. Here are some key points to remember:

• Rectilinear translation is the motion of an object along a straight line.

• To describe rectilinear translation, we need to use position, velocity and acceleration as functions of time.

• We can use four equations of motion for constant acceleration to solve rectilinear translation problems.

• We can also use position-time, velocity-time and acceleration-time graphs to represent and analyze rectilinear translation.

• Free fall is a special case of rectilinear translation where an object falls under the influence of gravity only.

• In free fall, an object accelerates downward at a constant rate equal to the acceleration due to gravity (g).

• We can apply the same equations of motion and graphs for free fall as for rectilinear translation, with the variables y and g.

FAQs

• What is the difference between displacement and distance?

Displacement is the net change in position of an object during its motion. It is a vector quantity, which means it has both magnitude and direction. Distance is the length of the path follo