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If you have constant acceleration, then you can't have constant velocity. (Unless the acceleration is constantly zero.)Final velocity = [initial velocity] + [ (acceleration) x (time) ]

V= v0 + kt where k is the constant acceleration and v0 is the initial velocity.

If, as you say, its acceleration is "constant", then the average is exactly equal to that constant.

If you know the initial and final velocity you can determine the acceleration (Velocity final- Velocity initial)/time = acceleration This can also be seen by integrating the acceleration. In this case lets assume acceleration is constant, then: acceleration=C Integration from time=initial to time=final gives C*(time final-time initial)=velocity final-velocity initial This integration scheme can also work if acceleration is not constant. In this case you must know how acceleration or velocity changes with time.

I would guess what you mean by a constant speed is zero change in acceleration. In the equation a = (v - u)/t where a = acceleration, v = final velocity, u = initial velocity and t = time taken for this change to happen. If your initial speed and final speed are the same (I.E. constant) then there is no acceleration.

Time = Speed/Accleration only if acceleration is constant, and initial velocity is 0.

velocity = distance / time There are also some formulae involving acceleration; for example, in the case of constant acceleration: velocity = initial velocity + acceleration x time If the acceleration is not constant, an integral is used instead.

If you have a particle with constant acceleration, and you add the initial and final velocities and then divide them by two, what you get is the average velocity of the particle in that period of time.

There are probably various formulae that involve initial velocity. For example: vf = vi + at, which applies for the case of constant acceleration.

Yes, it can, if the initial velocity vector of an object was in opposite direction to its constant acceleration. Example: Anything you toss with your hand has constant acceleration after you toss it ... the acceleration of gravity, directed downward. If you toss it upward, it starts out with upward velocity, which reverses and eventually becomes downward velocity.

Angular acceleration is got by the expression alpha = {(final angular velocity)2 - (initial ang velocity)2} / 2 theta. final is 50 and initial is 100 rad/s. Theta is 50 x 2pi radian Therefore required alpha = -50 x 150/200 pi = -75/2pi radian/s2 Negative sign indicates that the rotation is decelrated.

Is this a question? or a statement that you are unsure of? Well anyways, this would be correct if acceleration was a constant but if acceleration is not a constant, the (not-constant) acceleration would change the rate of velocity and thus that statement/question would be false.

If acceleration is constant over a period of time, then it is calculated as the change in velocity per unit time. If A is the acceleration over time T, U and V are the initial and final velocities respectively, thenA = (V - U)/TIf acceleration is not constant, then it is the derivative of the velocity with respect to time.

after an initial acceleration period,the elevator continues to move up with a constant speed.

You can use the formula for distance covered:distance = (initial velocity) x (time) + (1/2) (acceleration) (time squared) Solve for time. This assumes constant acceleration, by the way. If you assume that the initial velocity is zero, then you can omit the first term on the right. This makes the equation especially easy to solve.

zero because the initial and final velocity is constant . so,difference bet. final velocity and initial velocity is zero

The amount of time it would take an object to travel a distance with constant acceleration depends on its initial velocity, according to the equation: d = vit + 0.5at2 Where d is displacement, vi is initial velocity, t is time, and a is acceleration. Note: if the object starts from rest, its initial velocity, logically, is zero.

One formula that can be used - assuming constant acceleration, of course! - is vf2 = vi2 + 2as, where vf is the final speed, vi is the initial speed, a is the acceleration and s is the distance. In your case, solve for final velocity.

Wf - Wi = a*t, where Wi and Wf are the initial and final angular velocities, respectively, a is the angular acceleration, and t is time. So, a*t = 15.4 rad/s - 8.5 rad/s = 6.9 rad/s, thus a = 6.9 rad/s / 5.2 s = 1.3 rad/s2.

The formula is: V(T)=V0+at Wher V0=initial velocity a=acceleration in meters per second t=time

By integrating over time. Or for first year physics: s=at (constant acceleration; initial speed of zero) Hmm, distance = acceleration X time is how you read s=at.

Acceleration = Change in Speed divided by Time over which the change takes place. If acceleration is constant then Acc = [Final Speed - Initial speed] / Time If not, Acc = gradient of Speed-Time graph.

Assuming constant acceleration: distance = v(0) t + (1/2) a t squared Where v(0) is the initial velocity.

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