Theorem 1: For constant acceleration, the average velocity over any interval Ar

Question

Theorem 1: For constant acceleration, the average velocity over any interval Ar is equal to the mean of the initial and final velocities for that interval. That is, if the in acceleration is constant, iematic e Ax At v vo Vavg =- where v v() - the velocity at the end of the interval r, and vo e velocity at the beginning of the interval Ar Theorem 2: For constant acceleration, the average velocity over any interval ? is equal to the instantaneous velocity at the temporal midpoint. So, for constant acceleration, For the last expression, which involves "v" as a function of "t", we assume ?1-t-t.-t-0-t , so the temporal midpoint occurs at the time "t/2" 2 provides the crucial connection we need: it relates the average velocity (which we Theorem can actually measure in the lab with photogates) to an instantaneous velocity (which can easily be converted to the instantaneous speed needed to calculate the kinetic e absolute value). The first theorem is still important, however, because it is used to prove the second. You will be asked to prove both theorems at the end of this lab exercise. nergy by taking the

Explanation / Answer

Theorem 1.

ans) Velocity refers to rate of change of displacement with respect to time.

Its units: m/sec.

V= Distance/Time

In the theorem, it is stated that the acceleration is constant. Let the distance travelled is a, in a time t, from kinematics:

S= ut +(at2)/2

Average velocity V= s/t

Vavg = V0 + at/2

But acceleration*time = Change in velocity

so, at = V-V0

Vavg = V0+ (V-V0)/2

=(V+V0)2

So, velocity is caluculated as average velocity between the two points after a time interval dt secs

Instantaneous velocity is defined as the rate of velocity at which object is moving at a moment of time.

Since, the acceleration is specified as constant, Instantaneous velocity at any point is equal to the normal velocity.

Vins = V = change in diaplacement/time = dx/dt.

Hence, at constant acceleration, Vavg = Vins = dx/dt = (V+V0)/2.

Theorem 2:

ans) Given acceleration is constant, say a.

Integration of accelaration gives velocity at any time t.

so, V= at + c, c is a constant of integration.

Given time intervals are initial time interval t0 and final time interval t.

Their respective velocities are v(t0) = at0 + c

v(t) = at +c

Vavg = (v(t0) +v(t) )/2

= (at0 + c +at +c)/2

as at is rate of change of velocity

=V(t0 + t)/2

as initial velocity t0 =0.

Vavg = V(t/2) = Vins (at constant acceleration, instantanoeus velocity is equal to the velocity at temporal mid point)

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