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Question:WHAT IS THE FORMULA FOR DISPLACEMENT? I have to find the displacements for the following values.
I have a graph of velocity versus time for motion with variable acceleration. All that is given is the time in seconds. I have to obtain the velocity, acceleration and displacement and here is what I have gotten so far. Maybe you guys can draw this table it would help u understand my question better I cant make a table but heres the closest I got.
Time(s)Velocity (m/s)Acceleration (m/s^2)Displacement (m)
0.01.0
1.02.500.45?
2.03.250.31?
3.03.750.15625?
4.04.00.14?
5.03.500.25?
6.02.500.50?
7.02.00.75?
8.02.250.916?
9.03.751.125?
10.05.501.50?
the question marks are for the displacemnt which i dont know how to get.
DO THESE VALUES MAKE SENSE? Please give me formulas and a few examples as to how u got the answer for each value. Thanx!
Answers:Integrate the velocities using the trapezoid method to find the displacements: http://en.wikipedia.org/wiki/Trapezoidal_rule
Answers:Integrate the velocities using the trapezoid method to find the displacements: http://en.wikipedia.org/wiki/Trapezoidal_rule
Question:It seems as if I've forgotten this formula (we must have found it in my physics class though). I have a starting acceleration, a final acceleration, a displacement, but no time. I need to find an acceleration.. What's the equation?
Answers:I should think that if you have an initial and a final accel, and you need to find an accel, that you simply subtract a = a  a Now, this accel could be effected a number of ways: 1) a force (like gravity or friction) acting on a mass, a = F/m 2) a spring acting on a mass, a = kx/m 3) electromagnetic forces 4) etc. EDIT: I suspect that you meant initial and final VELOCITY, and a displacement. That's tricky. I don't know of such a formula. If there is one, I would think that it would have to be something like a = kv /x, where k is some dimensionless constant. v /x will give you the dimensions of acceleration. Or t = k(x/v) will yield seconds. Maybe t = x / v ? Try graphing velocity on the y axis and t on the x axis. (I know we don't know t, but some time has elapsed; don't use units.) If we assume UNIFORM acceleration, the line from (0 , v ) to (t , v ) will be a straight line. The area under the curve will be displacement. Maybe you can figure out how to make the area match your known displacement; that will give you t = t and you can go from there. That's all I've got. EDIT #2: I've been thinking about this, at the expense of some sleep. The graph described above i basically a rectangle with a right triangle on top (I assumed linear acceleration). The area is readily given as s = (v + v )t, so t = 2s / (v + v ) For uniform acceleration, a = v / t = (v  v )/t = (v  v )(v + v ) / 2s your formula! a < 0 if v < v Voila! FINAL EDIT: I was working another problem and I stumbled upon a more direct answer to your question. If you multiply the numerator terms in the eqn above, you get a = ( (v )  (v ) ) / 2s which can be rearranged as (v ) = (v ) + 2as which is Torricelli's equation.
Answers:I should think that if you have an initial and a final accel, and you need to find an accel, that you simply subtract a = a  a Now, this accel could be effected a number of ways: 1) a force (like gravity or friction) acting on a mass, a = F/m 2) a spring acting on a mass, a = kx/m 3) electromagnetic forces 4) etc. EDIT: I suspect that you meant initial and final VELOCITY, and a displacement. That's tricky. I don't know of such a formula. If there is one, I would think that it would have to be something like a = kv /x, where k is some dimensionless constant. v /x will give you the dimensions of acceleration. Or t = k(x/v) will yield seconds. Maybe t = x / v ? Try graphing velocity on the y axis and t on the x axis. (I know we don't know t, but some time has elapsed; don't use units.) If we assume UNIFORM acceleration, the line from (0 , v ) to (t , v ) will be a straight line. The area under the curve will be displacement. Maybe you can figure out how to make the area match your known displacement; that will give you t = t and you can go from there. That's all I've got. EDIT #2: I've been thinking about this, at the expense of some sleep. The graph described above i basically a rectangle with a right triangle on top (I assumed linear acceleration). The area is readily given as s = (v + v )t, so t = 2s / (v + v ) For uniform acceleration, a = v / t = (v  v )/t = (v  v )(v + v ) / 2s your formula! a < 0 if v < v Voila! FINAL EDIT: I was working another problem and I stumbled upon a more direct answer to your question. If you multiply the numerator terms in the eqn above, you get a = ( (v )  (v ) ) / 2s which can be rearranged as (v ) = (v ) + 2as which is Torricelli's equation.
Question:What's the formula to get the total displacement?
(a) Start from rest with constant accelera
tion of 2.45 m/s2 for 17.3 s;
(b) Constant velocity for the next
0.895 min;
(c) Constant negative acceleration of
8.86 m/s2 for 4.47 s.
What was the total displacement x for the
complete trip? Answer in units of m.
Answers:For (a) s1 = 0.5*a1*t1 let Vi be the velocity at the end of part (a); Vi = a1*t1 for (b) s2 = Vi*t2 = a1*t1*t2 for (c) s3 = Vi*t3  0.5*a3*t3 Total is s1 + s2 + s3
Answers:For (a) s1 = 0.5*a1*t1 let Vi be the velocity at the end of part (a); Vi = a1*t1 for (b) s2 = Vi*t2 = a1*t1*t2 for (c) s3 = Vi*t3  0.5*a3*t3 Total is s1 + s2 + s3
Question:a cross country runner runs 1,000 m north and then turns and runs 2,300 m in a direction 35 degrees west of north. I'm so confused. Please help!
Answers:Directions: + is East :::::::: + is North
Run in direction = 1000
Run in direction = 2300 ( u + v )
We know that 35 degrees west of north is 125 degrees counterclockwise from the East direction.
Let u + v be a unit vector. Since its length is 1, we can apply the rule that + = r = 1 and substitute u for and v for in the direction cosine relations:
cos = x/r = cos 125 =
sin = y/r = sin 125 =
and the unit vector is then (cos 125 ) + (sin 125 )
The vector we are looking for is then
2300(cos 125 ) + 2300(sin 125 ) = 1319.2258 + 1884.0497
and with the directions for and defined at the start,
we can say that the runner travels:
1319 m Westward
1884 m Northward, plus the additional 1000m before turning
I'm not adding these vectors ( 1319 ) + ( (1884+1000))
because the result wouldn't be the path he traveled.
Answers:Directions: + is East :::::::: +
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