# Homework Help: I need Help Plz can someone Help

1. Oct 1, 2004

### Tanya Back

I need Help :( Plz can someone Help

Hi i need help with motion problems..i am sooo confused..Can someone pLZZ help me with this question-->

A ladybug is at rest and accelerates to 2.0 m/s^2 when it sees a mite, this mite is 20.0 cm away from the ladybug and the mite is travelling at the constant speed of 5.00mm/s. How will long wit it take for the lady bug to reach the mite?

THANK U SOOOOOOOOO MUCH

2. Oct 1, 2004

### marlon

Ahh this is a classic one...

use the formula's for one-dimensional movements.
Suppose the bug starts moving at the origin of our reference frame (you can choose this as you want thanks to the genius of Descartes and Newton...)
Then the mite is 0.2m in front of the bug. so we get:

bug : x = 2tÂ²/2
mite : x = 0.2 + 0.005t (the velocity needs to be in m/s and it is given in mm/s !!!)

Equal these two equations and solve for t.

marlon

ps : normally i am not allowed to give you this much info but since you are new here I welcome you and this is your wedding-gift...enjoy and welcome to the jungle...

3. Oct 1, 2004

### Gokul43201

Staff Emeritus

Tanya, the policy here regarding homework is that you should show us that you've tried something, and are stuck. Barring this one time, we won't help with homework unless you've made an effort yourself. In the long run, you'll see that this is best for you.

Last edited: Oct 1, 2004
4. Oct 1, 2004

### cdhotfire

Good point gokul. Marlon should be spanked, .

5. Oct 3, 2004

### Tanya Back

Thank U!

Hey Thank u soo much Marlon...i kept on assuming dat distance was the same becasue when the lady bug has reached the mite they will be standing the same spot. To me dat meant the same distance :grumpy: anyways thank u verry much and I am sorry dat i didn't show my work before..i didn't know but next time i sure will. Thanks Again!!

Tanya

P.s - I totally agree with u Gokul43201...it will help me in the long run