# What is happening in this step, kinematics problem

• wahaj
In summary, the given problem asks for the time when a particle traveling along a straight line with an acceleration of 30 - 0.2v has a velocity of 30. The solution involves using integrals and the logarithmic property to find the time, but there is missing information about the initial velocity that affects the validity of the solution.
wahaj

## Homework Statement

a particle travels along a straight line and the acceleration is given by
a = 30 - 0.2v
determine the time when the velocity of the particle is v = 30

## Homework Equations

$$\int \frac {dv}{ax+b} = \frac{1}{a} \ln (ax+b)$$

## The Attempt at a Solution

This is the solution from the textbook. The problem is that my math class is behind schedule so I haven't learned integrals yet. The only knowledge I have on the matter is from what was briefly covered in my dynamics class. So I need an explanation of what is happening is happening in this solution.
$$dt = \frac{dv}{a}$$
$$\int ^t_0 dt = \int ^v_0 \frac {dv}{30 - 0.2v}$$
$$t\mid^t_0 = \frac {-1}{0.2} \ln({30-0.2v}) \mid ^v_0$$
I understand everything upto this point but I have no idea what is happening in the next step.
$$t = 5 \ln \frac {30}{30-0.2v}$$
$$t = 5 \ln \frac{30}{30 - 0.2(50)} = 1.12s$$
several questions, first where did the 30 in the numerator come from? why did the minus sign before $$\frac{-1}{0.2} = -5$$ go to and why was the v replaced by 50 instead of 30. There is no conversion of units going on in this question.

wahaj said:

## Homework Statement

a particle travels along a straight line and the acceleration is given by
a = 30 - 0.2v
determine the time when the velocity of the particle is v = 30

## Homework Equations

$$\int \frac {dv}{ax+b} = \frac{1}{a} \ln (ax+b)$$
Typo above - it should be dx in the first integral.
wahaj said:

## The Attempt at a Solution

This is the solution from the textbook. The problem is that my math class is behind schedule so I haven't learned integrals yet. The only knowledge I have on the matter is from what was briefly covered in my dynamics class. So I need an explanation of what is happening is happening in this solution.
$$dt = \frac{dv}{a}$$
$$\int ^t_0 dt = \int ^v_0 \frac {dv}{30 - 0.2v}$$
$$t\mid^t_0 = \frac {-1}{0.2} \ln({30-0.2v}) \mid ^v_0$$
I understand everything upto this point but I have no idea what is happening in the next step.
The right side above is -5[ln(30 - .2v) - ln(30)] = 5[ln(30) - ln(30 - .2v)]
$$= 5 ln\frac{30}{30 - .2v}$$
They're just using a property of logs; namely that ln(a/b) = ln(a) - ln(b).
wahaj said:
$$t = 5 \ln \frac {30}{30-0.2v}$$
$$t = 5 \ln \frac{30}{30 - 0.2(50)} = 1.12s$$
several questions, first where did the 30 in the numerator come from? why did the minus sign before $$\frac{-1}{0.2} = -5$$ go to and why was the v replaced by 50 instead of 30. There is no conversion of units going on in this question.

did your post cut off or something?

wahaj said:
did your post cut off or something?

No, I think it's all there. Look at the part before the last quote. That's answering the question of why 5 instead of -5 and the manipulation of the logs stuff.

well i managed to figure it out when I found out that the log property was used. I don't understand the point of putting the variable in the denominator instead of simply writing
$$\ln \frac{30-0.2v}{30}$$
why make things unnecessarily complicated just to get rid of a negative sign.
Anyways thanks for the help.

wahaj said:
well i managed to figure it out when I found out that the log property was used. I don't understand the point of putting the variable in the denominator instead of simply writing
$$\ln \frac{30-0.2v}{30}$$
why make things unnecessarily complicated just to get rid of a negative sign.
Anyways thanks for the help.

Yes, there was no NEED to get rid of the negative. They just chose to write it that way.

wahaj said:

## Homework Statement

a particle travels along a straight line and the acceleration is given by
a = 30 - 0.2v
determine the time when the velocity of the particle is v = 30

## Homework Equations

$$\int \frac {dv}{ax+b} = \frac{1}{a} \ln (ax+b)$$

## The Attempt at a Solution

This is the solution from the textbook. The problem is that my math class is behind schedule so I haven't learned integrals yet. The only knowledge I have on the matter is from what was briefly covered in my dynamics class. So I need an explanation of what is happening is happening in this solution.
$$dt = \frac{dv}{a}$$
$$\int ^t_0 dt = \int ^v_0 \frac {dv}{30 - 0.2v}$$
$$t\mid^t_0 = \frac {-1}{0.2} \ln({30-0.2v}) \mid ^v_0$$
I understand everything upto this point but I have no idea what is happening in the next step.
$$t = 5 \ln \frac {30}{30-0.2v}$$
$$t = 5 \ln \frac{30}{30 - 0.2(50)} = 1.12s$$
several questions, first where did the 30 in the numerator come from? why did the minus sign before $$\frac{-1}{0.2} = -5$$ go to and why was the v replaced by 50 instead of 30. There is no conversion of units going on in this question.

There is some vital information missing from the problem statement: what is the initial velocity v0 at t = 0? For example, if v0 = 50 we cannot have v(t) = 30 for t > 0; the solution has t < 0; if v0 > 150 there is no time where v(t) = 30.

## 1. What is kinematics?

Kinematics is the branch of physics that studies the motion of objects without considering the forces that cause the motion. It involves analyzing the position, velocity, and acceleration of objects as they move through space and time.

## 2. What is a kinematics problem?

A kinematics problem is a type of physics problem that involves using equations and principles of motion to solve for unknown quantities, such as position, velocity, or acceleration, of an object.

## 3. How do I approach a kinematics problem?

To approach a kinematics problem, you should first identify the known and unknown quantities, then choose the appropriate equations to use based on the given information. It is also important to draw a diagram to visualize the problem and label all the variables.

## 4. What are the common equations used in kinematics problems?

The most commonly used equations in kinematics problems are the equations of motion, which include distance formula, velocity formula, and acceleration formula. Other important equations include the equations for uniform motion and motion with constant acceleration.

## 5. How can I check if my answer to a kinematics problem is correct?

To check if your answer is correct, you can plug your values back into the original equations and see if they satisfy the given conditions. It is also helpful to double-check your units and make sure they are consistent throughout the problem.

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