This should be really simply but I can't figure it out... the problem is: A car is driving along at 11 m/s, and the rain is falling at 3 m/s straight down. What angle q (in degrees) does the rain make with respect to the horizontal as observed by the driver
If someone could even tell me how to set this up it would be a great help... i'm completely lost
2006-09-08 08:59:50 · 9 answers · asked by blue_angel_1400 2 in Science & Mathematics ➔ Physics
Relative to the car, the rain is traveling 11 m/s backwards and 3 m/s down. Add the two vectors together (11 units left, 3 units down) and draw a line from the beginning of the first vector to the end of the second. Use trig (tangent) to solve for the angle.
tan a = 3/11
a = atan 3/11
2006-09-08 09:11:05 · answer #1 · answered by Automation Wizard 6 · 0⤊ 0⤋
It is just a right triangle. The two sides making the right angle are 11 for the velocity of the car and 3 for the velocity of the rain. The hypotenuse if the velocity (speed and direction) of the rain hitting the car. Solve for the angles. Since the car is driving faster than the rain is falling, the larger of the two angles is the one you want.
2006-09-08 09:10:21 · answer #2 · answered by campbelp2002 7 · 0⤊ 0⤋
Draw a triangle with a base 11 units long and one side at right angles to it 3 units high. The length of the third side will give you the apparent speed of the rain and it's angle to the base the angle of the rain.
2006-09-08 11:24:01 · answer #3 · answered by Stewart H 4 · 0⤊ 0⤋
it's so easy. what you should know is a little about relativity.
when the car is driving at 11m/s and you are in the car. it's like that the world is coming towards you at 11m/s. so you can see that rain is falling down at 3m/s & meanwhile, it's coming towards you at 11m/s. so it has a vector of (-11,-3) m/s.
the angle of that vector is simply arc-tangent(11/3)
2006-09-08 09:10:45 · answer #4 · answered by Sally 3 · 0⤊ 1⤋
arctan(3/11)=15degrees
Just draw a figure with the two speeds as a triangle.
2006-09-08 09:07:14 · answer #5 · answered by Anonymous · 1⤊ 0⤋
you're merely getting perplexed with all those deltas and stuff. Do one ingredient. evaluate the equation to be like this : av. speed = distance / time. Now, they merely gave distance as delta x and time as delta t. All you should do is, delta t = delta x / Vav. = 138 / 40 9.2 = 2.80 hours it really is the authentic of that.
2016-10-15 23:38:20 · answer #6 · answered by ? 4 · 0⤊ 0⤋
Use vectors, find the components, ill set this up for you
Vx1 = 11cos0
Vy1 = 11sin0
Vx2 = 3cos90
vy2 = 3sin 90
there its set up
2006-09-08 09:05:26 · answer #7 · answered by sur2124 4 · 0⤊ 1⤋
It is a relative velocity question. Va=Vb+Va/b. Velocity is a vector quantity.
2006-09-08 11:08:07 · answer #8 · answered by daedgewood 4 · 0⤊ 0⤋
it makes two points
2006-09-08 09:05:14 · answer #9 · answered by jaydaka 2 · 0⤊ 2⤋