# Homework Help: Kinematic question

1. Apr 21, 2010

### gaobo9109

1. The problem statement, all variables and given/known data
A particle of mass m is projected vertically upwards, with an initial speedo of u, in a medium which exerts a resistance of magnitude mkv, where k is a positive constant and v is the speed of the particle after time t. Express, in terms of k, g and u, the time taken for the particle to reach its greatest height

2. Relevant equations
u - gt = 0

3. The attempt at a solution
u - gt - kvt = 0
I don't know how to express v in terms of u, g and k.

2. Apr 21, 2010

### kuruman

Can you find an expression for the acceleration of the particle?

3. Apr 22, 2010

### gaobo9109

a = g + kv?

4. Apr 22, 2010

### kuruman

Basically yes, but I would put a negative sign in front of both terms. Now can you write an expression for the rate of change of velocity with respect to time?

5. Apr 22, 2010

### gaobo9109

Isn't dv/dt = a = -g - kv?

6. Apr 22, 2010

### kuruman

Correct. Now can you solve this equation to find v(t)?

7. Apr 22, 2010

### gaobo9109

dv/dt = -g - kv
dv = -g - kvdt
Integrating both side
v = -gt - kvt
v = -gt / (1 + kt)

I just learnt about differential equation, I am not sure if this is correct

8. Apr 22, 2010

### kuruman

It is not correct. You need to separate the two variables before you integrate. This means that you need to rearrange the equation algebraically so that the left side has only v (and constants) in it and the right side has only t (and constants) in it.

9. Apr 22, 2010

### gaobo9109

Sorry, I am quite new in differential equation. I don't know how to rearrange the equation such that v is on one side and t on the other side. Can you please teach me? Thanks

10. Apr 22, 2010

### kuruman

This part of the process is "algebra" not "differential equations".

$$\frac{dv}{dt}=-(g+kv)$$

Cross multiply to get

$$\frac{dv}{g+kv}=-dt$$

Voila. The variables have been separated. Now you can integrate. Differential equations are not as scary as they sound. Now you can integrate.