1.)For the velocity vectors shown determine d) the direction of the vector D. 2.
ID: 1692120 • Letter: 1
Question
1.)For the velocity vectors shown determine d) the direction of the vector D. 2.) A car traveling at 35mi/h is to stop on a 35m long shoulder of the road.a) Convert velocity:
b) What is the required magnitude of the minimum acceleration?
c) How much time will elapse during the minimum deceleration until the car stops? 3.)Assume that a motorcycle left the take-off ramp at 120 to the horizontal and that the take-off and landing heights are the same. It went 77m. Neglect air resistance.
a) What was his take off speed? b)How long he has been in the air? c)What was his maximum height in the air?
A car traveling at 35mi/h is to stop on a 35m long shoulder of the road.
a) Convert velocity:
b) What is the required magnitude of the minimum acceleration?
c) How much time will elapse during the minimum deceleration until the car stops? 3.)Assume that a motorcycle left the take-off ramp at 120 to the horizontal and that the take-off and landing heights are the same. It went 77m. Neglect air resistance.
a) What was his take off speed? b)How long he has been in the air? c)What was his maximum height in the air?
Explanation / Answer
1. Lets first find the components of each of the vectors and place them into component notation to make answering the questions in problem 1 a little easier. A = 5i + 0j = <5,0> --> Since vector A has no vertical component, it is simply its magnitude in the positive x directionB = 5i + 8.66j = <5,8.66> --> Since vector B has a vertical component and we are given the angle, we can easily figure out the components
of this vector. to find the vertical component, you need
to take the sin of the angle and multiply it by the magnitude of the vector. so in this case it is sin(60)*10 = 8.66 in the y direction. the same goes for the horizontal component = cos(60)*10 = 5
C = -12.99i + 7.5j = <-12.99,7.5> --> This one is a little more tricky. the thing to remember is the x component for this vector
is negative. simply take the sin and cos times the magnitude for each component but remember to make the x component negative. now we can answer those questions much easier.
a. we just found these, they are as follows: Ax= 5 Bx= 5 Cx = -12.99
b. we also just found these, they are: Ay = 0 By = 8.66 Cy = 7.5 c. to add vectors, simply add their respective components. Thus the x component of D = 5 + 5 - 12.99 = -2.99 and the y component of D = 0 + 8.66 + 7.5 = 16.16 so the answer is <-2.99,16.16> or -2.99i + 16.16j
d. this is another trig problem. we have the components (the sides of a triangle) so we can find the angle by taking the arctan of them. tan-1(16.16/-2.99) = -79.51o this is the angle starting from the negative x axis and moving clockwise.
2.
a. I assume by converting, you simply mean mi/h to m/s. first, we can convert mi/h to mi/s by dividing the speed by 3600s/h which gives us .009722mi/s and then to m/s by multiplying by 1609.344m/mi which gives us 15.65m/s.
b. this can be done with a kinematic equation vf2 = vi2 + 2ad where d is the distance traveled and a is the acceleration. the distance is 35m and the final velocity is 0 so the equation looks like: 0 = 15.652 + 70a
through some simple algebra we can find that a = -3.49m/s2
c. also another kinematic using the equation v = at. v = -15.65 a = -3.49
through evaluating it we find t = 4.48s 3. a. use the range equation which looks like R = v2sin(2?)/g where ? = the launch angle and g = the acceleration due to gravity and R is the horizontal distance traveled. so set up it looks like 77m = v2sin(24)/9.8m/s2
again by evaluating the equation we get that v = 43.07m/s '
b. take the time for him to reach 0m/s vertically and multiply by 2. so v = at. his initial vertical velocity was sin(12)*43.07= 8.95m/s so if we plug that into the equation we get that t = 1.83s
c. use the max height equation, which looks like y = (vsin(?))2/2g where this v is the total velocity. again plug it in and you get 4.1m
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