Help with physics please!?

Question

Three cars are moving at constant speeds relative to the ground. One heading due east, the other is heading due north and the third is heading due west. It is found that the north-bound car is moving with a speed of: (1) 32 mph as measured by the ground, (2) 40 mph as measured by the west-bound car and (3) 68 mph as measured by the east-bound car. Determine the speed of the east-bound caras measured by the west-bound car.

Now, I know that I need to write out the velocities as vectors and somehow add them or use the distance formula to solve it. I'm just not sure how to write the vectors out in i + j form or what to do after that is done.

Thanks to those who help!

Answer 1

True that, use the i+j format. Lets say east is positive i and north is positive j. Then the cars are moving as follows:

north bound car = 32j

east bound car = (a)i

west bound car = (b)i   (b will be negative)

(2) moving 40 mph according to west-bound, thus

|32j + (b)i| = 40

Yip, here you use pythagoras, what must b be so that:

32^2+b^2 = 40^2

b = +- root of (576)

b = +- 24 but as we chose the east as positive i, then b must be the negative, thus

b=-24

(3) here the same goes for east bound

|32j+(a)i|=68

again, pythagoras, try to get: 60

a=60

Now, the east bound car against the west bound car:

the two vectors of the cars are

car east = (60)i

car west = (-24)i

But remember, you want to measure in relation to each other, the two vectors are point away from each other.

Total speed of east bound in relation to west bound is

Total = (60)i-(-24)i

        = (84)i

        84mph

(Best answer please

Answer 2

You can just use Pythagoras' theorem on this one.

32^2+W^2=40^2

32^2+E^2=68^2