In uniform circular motion, what are the average velocity and average acceleration for one revolution? Explai?
They are both zero. Remember displacement, velocity and acceleration are vectors. Direction is important. E.g. you walk 2m left and 2m right, your displacement is zero, because you have ended up where your started from. Velocity first. This equals displacement/time. For 1 full revolution, your displacement is zero, as described above. So your average velocity is zero. You have basically cancelled out velocity in any given direction by the reverse velocity, coming back to your starting point. Acceleration is similar. a=velocity_change/t. Your velocity is the same at the start and at the end of a full revolution. So a= 0. If you want to think about it a different way, acceleration is the centripetal acceleration, magnitude mv^2/r. The direction is continuously changing; it acts along a radius towards the centre, and this radius is constantly rotating. If the circle was horizontal for example, the acceleration would spend as much time acting left as it acts to the right. The net effect is zero average acceleration.
Finding uniform acceleration?!?
This can be solved using one of the kinematic equations of motion. We have the following information: u = initial speed = 32 m/s v = final speed = 96 m/s t = time = 8.0 s a = acceleration = ? v = u + at. Subtracting u from both sides, at = (v-u). Dividing through by t, a = (c-u)/t = (96-32)/8 = 8.0 m/s².
How can you determine the formula for acceleration?
I guess from the question that the answers offered are probably a bit too complex.So, starting from the beginning:A stationary object has fixed co-ordinates (in a given frame of reference), it’s position.A moving object changes its position over time, i.e. it moves from one set of co-ordinates to a different set in a given interval of time.The seperation between the two positions (determined geometrically) is denoted as ‘S’.The time interval is denoted as ‘T’The speed (the rate of change of position with time) is given by S/T.The velocity is numerically the same as the speed but takes into account the direction of the change of position within the frame of reference but is still S/T.When an object at rest begins to move there is a change in it’s velocity (from 0 to some other value) over some time interval. This is called acceleration and denoted as ‘A’. The same applies if there is any change in V.The rate of change in velocity is V/TTherefore:A = V/T and V = S/TSo A = S/T^2Hope that helps
What is the formula for average velocity?
Q: What is the formula for average velocity?A: It turns out this question contains a land-mine. If you have done the kinematics equations you KNOW that average velocity is start velocity plus end velocity divided by two. In general this is wrong however. It holds in the restricted case of CONSTANT ACCELERATION. This almost never happens in real life. A better rule is to remember “Don’t Average Ratios”. Velocity is a ratio, of distance over time. If you add the numerators and the denominators - the top and bottom of each ratio, you GUARANTEE the correctness of the final ratio.I drive at 20 mph for 1/2 an hour and 40 mph for 1 hour. Average speed?NOT (20 + 40) / 2 = 30 mph Wrong!Instead total distance / total time =(10 + 40) / ( 1/2 + 1) = 50 / 1.5 = 33.3 mph Right!
How do I find velocity without acceleration?
We can use the kinematic equation: d = d₀ + (1/2)(v + v₀)t. Solving for v gives: v = 2(d - d₀)/t - v₀. We have that: d = 207.61 m d₀ = 0.00 m t = 9.77 s v₀ = 0.00 m/s. Substituting in our values yields the velocity to be: v = 2(d - d₀)/t - v₀ = 2(207.61 m - 0.00 m)/(9.77 s) - 0.00 m/s = 42.5 m/s. I hope this helps!
Average acceleration in circular motion?
note that for uniform rotational movement, accceleration always points to center of the rotation. so movement from 6 to 3 o'clock will have average that points north west (halfway between 12 and 9 o clock). now let's find magnitude of it: first find average velocity for travel between 6 and 3 o'clock : vavg=v2-v1 since magnitude of v1 and v2 is 4m/s but they are 90 deg appart, (one pointing east, other pointing north) using basic trigonometry or pytaghorean: vavg=sqrt(2)*4m/s vavg=5.66m/s (this is one part, now we need time) time to cover 1/4 of a circle is then t=0.25*(2*pi*r)/v t=pi/v t=0.7854s then using averages aavg=vavg*t aavg=7.20m/s^2
How to find acceleration given velocity and distance?
You have used a wrong formula distnce/velocity cam never give you acceleration it will give you some quantity which hs dimensions of time Either use vf^2 -vi^2 = 2*a*s, or conventinally less frequently used formula disatnce / Average velocity = time and then use (vf-vi)/time = a I use second approch as it does not involve squares time =0.0221/[(3290000 + 45400)/2] = 1.3252*10^-8 a = (3290000 - 45400)/(1.3252*10^-8) = 2.45*10^14 m/s^2
How do you find final velocity from acceleration and distance?
The problem is not clearly stated, or terms well defined. Is this a problem of motion in one dimension? What distance? Do you mean by "distance" the length of the displacement from initial position to final position? If it is one-dimensional, is acceleration given to be constant, as assumed by the two answers given by Paul A Allcock and Austin Douglas? In any case, to determine the final velocity you must be given two initial conditions, usually the initial velocity and initial position. To use the one-dimensional constant acceleration formulas to find the final velocity, you need in addition either the final position, or the time it takes to reach the final velocity. Since you don't mention any times, you must be assuming that the initial position [math]x_i[/math] and final position [math]x_f[/math] are known, as well as the initial velocity [math]v_i[/math]. The relevant equation is then the one used by the other two answerer's, namely, [math]\displaystyle v_f^2 - v_i^2 = 2a\,|x_f - x_i|[/math], which can be solved by adding [math]v_i^2[/math] to both sides, then taking the square root of the result, to obtain an expression for the final velocity: [math]\displaystyle v_f = \sqrt{2a\,|x_f - x_i| + v_i^2}[/math]. The absolute value, or magnitude, of the displacement [math](x_f - x_i)[/math], is the distance traveled as the velocity changes from [math]v_i[/math] to [math]v_f[/math]. The initial velocity may, but need not, be zero. No value has been given for it in the problem statement. If the acceleration is in the opposite direction of the displacement [math](x_f - x_i)[/math], then [math]a[/math] should be replaced by [math]-a[/math].
What is the formula for maximum acceleration? How is this formula determined?
The formula for maximum acceleration usually has to be derived from other forces, as the actual change in velocity cannot happen at a particular instance, while the maximum acceleration can.A=F/M (Acceleration = Force/Mass) is a good calculation for acceleration, and is typically the one you’d use for maximum acceleration.However, if the problem exhibited changing velocity as a formula, or provided a range of velocities with no other information, then you could use the derivative of the function to work it out, or simply calculate over the quantisation period specified in the problem.Typically when calculating maximum acceleration ( in my case, with respect to weapon shock calculations which I have to do quite a bit ) I use the first, however at times I also need to take other factors into account, since mass isn’t always constant and varies with time according to shock wave propagation through a medium.In such circumstances, I then attempt to predict the propagation pattern and choose a quantization method for breaking up the solution in the absence of a formula to track it - that is, I might calculate it based on a centimeter, a meter or even a millimeter. This kind of calculation is more complicated and I still use F=MA, but need to consider smaller masses and smaller force changes over time as they propagate through the now-quantized object. A simulation is often the easiest way to calculate such results.
What is the formula to calculate vertical acceleration? How is this determined?
Acceleration is defined as the rate of change of velocity with respect to time. (Velocity being the rate of change of position with respect to time). So, we need to know something about the (vertical) forces acting on the object, seeing as by Newton’s second law, force is proportional to acceleration.If it’s a random object thrown near the earth’s surface, the main force determining the thing’s motion is gravity. It varies a fraction and gets weaker at greater distances from the center of mass of the system (middle of the Earth, in this case) but there is a well-known value:[math]9.8\frac{m}{s^2} \:(meters\:per\:second\:per\:second)[/math]for the acceleration due to gravity at or around Earth’s surface. Obviously, this points downward, not up!This is an overview. Newton’s second law is perhaps the next layer of detail to understand. Feel free to ask more.