# Homework Help: Magnitude - Distance

1. Oct 2, 2007

### willingtolearn

[SOLVED] Magnitude - Distance

#1
The force of the wind on the sails of a sailboat is 390 N north. The water exerts a force of 180 N east. If the boat (including crew) has a mass of 270 kg, what are the magnitude and direction of its acceleration.
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390 N North, 180 N East.
m = 270 kg

I never know how to find the magnitude, can some one help or explain the magnitude to me ?

#2
A shopper in a supermarket pushes a loaded cart with a horizontal force of 10 N. The cart has a mass of 30 kg. HOw far will it move in 3.0 s, starting from the rest? (Ignore friction)
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Horizontal F = 10 N
m = 30kg
t = 3 s
d = rt
v (initial) = 0

I only got to this point ?

2. Oct 2, 2007

### Kurdt

Staff Emeritus
Well for part one you simply use pythagoras' theorem. Think of the components of force as two sides of a right angled triangle. For part two you will need to use one of the kinematic equations. Are you familiar with these?

They can be found here: https://www.physicsforums.com/showthread.php?t=110015

3. Oct 2, 2007

### willingtolearn

#1
So the magnitude is the same as the resultant !
#2
I only know to the Newton 2nd Law

4. Oct 2, 2007

### Kurdt

Staff Emeritus
A resultant vector is a sum of vectors, the magnitude of the vector is the length of the vector.

For number two try applying the following kinematic equation.

$$s = ut+\frac{1}{2}at^2$$

,where s is the distance travelled and u is the initial velocity.

5. Oct 2, 2007

### willingtolearn

#1
mag = 429.5 N
acc is NorthEast with the value 1.6 m/s2

#2
OH, i forgot that equation ! Forgot to think of the term displacement !
d = 1.5 m
Is that correct ?

6. Oct 2, 2007

### Kurdt

Staff Emeritus
You may need to work out an angle for number 1. I'm not sure if they'll let you get away with just north east. Other than that you're correct.

7. Oct 2, 2007

IK ! Thanks