- #1

wahaj

- 156

- 2

## 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

[tex]\int \frac {dv}{ax+b} = \frac{1}{a} \ln (ax+b) [/tex]

## 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.

[tex] dt = \frac{dv}{a} [/tex]

[tex]\int ^t_0 dt = \int ^v_0 \frac {dv}{30 - 0.2v} [/tex]

[tex]t\mid^t_0 = \frac {-1}{0.2} \ln({30-0.2v}) \mid ^v_0 [/tex]

I understand everything upto this point but I have no idea what is happening in the next step.

[tex] t = 5 \ln \frac {30}{30-0.2v} [/tex]

[tex] t = 5 \ln \frac{30}{30 - 0.2(50)} = 1.12s [/tex]

several questions, first where did the 30 in the numerator come from? why did the minus sign before [tex] \frac{-1}{0.2} = -5 [/tex] go to and why was the v replaced by 50 instead of 30. There is no conversion of units going on in this question.