# A car travels 20 km due

1. Aug 22, 2010

### r-soy

Hi all

A car travels 20 km due north and then 35 km in a direction 60 degree west of north. Find the magnitude and direction of the car's resultant displacement.

How I can solve like this question ?

I try to solve :

the magnitude = R = A + B

= 55 km

direction of the car's resultant displacement

is north

2. Aug 22, 2010

### Mentallic

Do you understand what displacement is?

If I go 1km north and then 1km east, my final displacement is the distance (magnitude) and direction I am from my starting position. So the answer to this problem would be 45o East of North with a distance of $$\sqrt{2}$$km.

3. Aug 22, 2010

### r-soy

can I solve this question by this two rule for

magnitude = R = sqrt(A^2 + B^2)
dircaion = tan = y/x

????

4. Aug 22, 2010

### Mentallic

Yes but you have to think a little more about the problem to set it up into a right-triangle so you can use those formulas.

5. Aug 22, 2010

### HallsofIvy

Staff Emeritus
I would not do it as a right triangle problem. You have a triangle with two sides of lengths 20 and 35. Since the first leg is "due north" and the second leg is "60 degrees" west of north", the angle between the two legs has measure 180- 60= 120 degrees. Since you have two sides of the triangle and the angle between them, use the "cosine law": $c^2= a^2+ b^2- 2ab cos(C)$ where a and b are two sides of the triangle, C is the angle between them, and c is the third side, opposite C.

Here, $c^2= 20^2+ 35^2- 2(20)(35)cos(120)$

6. Aug 22, 2010

### Redbelly98

Staff Emeritus
Moderator's note: this thread has been moved from General Math to the Homework & Coursework Questions area of the forum.

7. Aug 24, 2010

### r-soy

thank you but c (third side ) mean magnitude ??

and now how to find the dircaion ??

8. Aug 25, 2010

### HallsofIvy

Staff Emeritus
Since you are given 2 sides of the triangle (and can calculate the third) and one angle, you could use the sine law to find the other angles: If A, B, and C are three angles in a triangle and a, b, and c are the lengths of the sides opposite each side, respectively, then
$$\frac{sin(A)}{a}= \frac{sin(B)}{b}= \frac{sin(C)}{c}$$