Kinematics And One-Dimensional Motion: Non-Constant .

Kinematics andOne-Dimensional Motion:Non-Constant Acceleration8.01W01D3

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Average Velocity The average velocity, v ave (t) , is the displacement Δrdivided by the time interval Δt v ave Δr Δx î vave,x (t) îΔt ΔtThe x-component of the average velocity is givenbyvave,xΔx Δt

Instantaneous Velocity andDifferentiationFor each time interval Δt , calculate the x-component ofthe average velocityvave,x (t) Δx / ΔtTake limit as Δt 0 sequence of the x-componentaverage velocitiesΔxx(t Δt) x(t) dxlim lim Δt 0 ΔtΔt 0ΔtdtThe limiting value of this sequence is x-component ofthe instantaneous velocity at the time t.vx (t) dx / dt

Instantaneous Velocityx-component of the velocity is equal to the slope of thetangent line of the graph of x-component of position vs.time at time tdxvx (t) dt

Worked Example: Differentiationx(t) At 2x(t Δt) A(t Δt)2 At 2 2AtΔt AΔt 2x(t Δt) x(t) 2At AΔtΔtdxx(t Δt) x(t) lim lim (2At AΔt) 2AtΔt 0Δt 0dtΔtGeneralization for Polynomials:x(t) At ndx nAt n 1dt

Concept Question: InstantaneousVelocityThe graph shows the position as afunction of time for two trainsrunning on parallel tracks. Fortimes greater than t 0, which ofthe following is true:1. At time tB, both trains have thesame velocity.2. Both trains speed up all the time.3. Both trains have the same velocityat some time before tB, .4. Somewhere on the graph, bothtrains have the same acceleration.

Average AccelerationChange in instantaneous velocity divided bythe time interval Δt t2 t1 a ave (vx,2 vx,1 )ΔvxΔv Δvx î î î aave,x îΔtΔtΔtΔtThe x-component of the average accelerationaave,xΔvx Δt

Instantaneous Accelerationand DifferentiationFor each time interval Δt , calculate the x-component ofthe average accelerationaave,x (t) Δvx / ΔtTake limit as Δt 0 sequence of the x-componentaverage accelerationsΔvxvx (t Δt) vx (t) dvxlim lim Δt 0 ΔtΔt 0ΔtdtThe limiting value of this sequence is x-component ofthe instantaneous acceleration at the time t.ax (t) dvx / dt

Instantaneous AccelerationThe x-component of acceleration is equal to the slope of thetangent line of the graph of the x-component of the velocityvs. time at time tdvxax (t) dt

Group Problem: Model RocketA person launches a home-built model rocket straight upinto the air at y 0 from rest at time t 0 . (The positive ydirection is upwards). The fuel burns out at t t0. Theposition of the rocket is given by 1a0 6 42y (a0 g)t t / t0 ;30 20 t t0with a0 and g are positive. Find the y-components of thevelocity and acceleration of the rocket as a function oftime. Graph ay vs t for 0 t t0.

Non-Constant Accelerationand Integration

Change in Velocity: Integral ofAccelerationConsider some time t such thatt0 t t fThen the change in the x-componentof the velocity is the integral of the xcomponent acceleration(denote vx,0 vx (t0 ) ).vx (t) vx,0 t ′ t ax (t ′ ) dt ′t ′ t0“Integration is the inverse operation of differentiation”

Change in Position: Integral ofVelocityArea under the graph of x-component of the velocity vs. time isthe displacement (denote x0 x(t0 )).x(t) x0 t ′ t t ′ t0vx (t ′ ) dt ′

Worked Example: Time-DependentAcceleration2a(t) AtAcceleration is a non-constant function of time xwith t0 0 , vx,0 0 , and x0 0 .Change in velocity:t ′ tt ′ t3tAt′vx (t) 0 At ′ 2 dt ′ A vx (t) 3 t′ 03t ′ 0Change in position:t ′ tt ′ t34 t′ t′ At 4x(t) 0 A dt ′ x(t) A 12 3 12 t ′ 030Generalization for Polynomials:t ′ tn 1 t ′ tn 1n 1ttt′n0tdt ′′ n 1 t ′ t n 1 n 1t ′ t 00

Special Case:Constant Accelerationax constantAcceleration:Velocity:vx (t) vx,0 t ′ t axdt ′ a x t t ′ 0vx (t) vx,0 ax tPosition:x(t) x0 t ′ t (vx,0 ax t ′ ) dt ′ t ′ t01x(t) x0 vx,0t ax t 22

Concept Question: IntegrationA particle, starting at rest at t 0, experiences a nonconstant acceleration ax(t) . It’s change of position can befound by1. Differentiating ax(t) twice.2. Integrating ax(t) twice.3. (1/2) ax(t) times t2.4. None of the above.5. Two of the above.

Group Problem: Sports CarAt t 0 , a sports car starting at rest at x 0 accelerateswith an x-component of acceleration given byax (t) At Bt , for 0 t (A / B)31/2and zero afterwards with A, B 0(1) Find expressions for the velocity and position vectors ofthe sports car as functions of time for t 0.(2) Sketch graphs of the x-component of the position, velocityand acceleration of the sports car as a function of time fort 0

Appendix:Integration and the RiemannSum

Change in Velocity: Area UnderCurve of Acceleration vs. timeMean Value Theorem: Foreach rectangle there exists atimeti t ci ti 1such thatdvxvx (ti 1 ) vx (ti ) (ci )Δt ax (ci )Δtdt

Apply Mean Value Theoremvx (t1 ) vx (t0 ) ax (c0 )Δt(vx (t2 ) vx (t1 )) ax (c1 )Δt vx (ti 1 ) vx (ti ) ax (ci )Δt vx (tn ) vx (tn 1 ) ax (cn 1 )ΔtWe can add up the area of the rectangles and findvx (tn ) vx (t0 ) i n 1 ((a (c )Δt)xi 0i

Change in Velocity: Integral ofAccelerationThe area under the graph ofthe x-component of theacceleration vs. time is thechange in velocityi Nvx (t f ) vx (t0 ) lim ax (ti )ΔtiΔti 0i 1t t fvx (t f ) vx (t0 ) ax (t) dtt zTGUlIASZx83u

Velocity The graph shows the position as a function of time for two trains running on parallel tracks. For times greater than t 0, which of the following is true: 1. At time t B, both trains have the same velocity. 2. Both trains speed up all the time. 3. Both trains have the sam