Kinematics and One-Dimensional Motion: Non-Constant Acceleration 8.01 W01D3 Announcements Familiarize Yourself with Website https://lms.mitx.mit.edu/courses/MITx/8.01/2014_Fall/about Buy or Download Textbook (Dourmashkin, Classical Mechanics: MIT 8.01 Course Notes Revised Edition ) Downloadable links (require certificates): https://lms.mitx.mit.edu/courses/MITx/8.01/2014_Fall/ courseware/Intro/about:Resources/ Buy Clicker at MIT Coop Sunday Tutoring in 26-152 from 1-5 pm Average Velocity The average velocity, v ( t ) , is the displacement Δ r ave divided by the time interval Δt Δr Δx ˆ ˆ vave ≡ = i = vave,x (t)i Δt Δt The x-component of the average velocity is given by Δx vave,x = Δt Instantaneous Velocity and Differentiation For each time interval Δ t , calculate the x-component of the average velocity v (t) = Δx / Δt ave,x Take limit as Δ t → 0 sequence of the x-component average velocities Δx x(t + Δt) − x(t) dx lim = lim ≡ Δt→0 Δt Δt→0 Δt dt The limiting value of this sequence is x-component of the instantaneous velocity at the time t. v (t) = dx / dt x Instantaneous Velocity x-component of the velocity is equal to the slope of the tangent line of the graph of x-component of position vs. time at time t dx vx (t) = dt Worked Example: Differentiation x(t) = At 2 x(t + Δt) = A(t + Δt)2 = At 2 + 2AtΔt + AΔt 2 x(t + Δt) − x(t) = 2At + AΔt Δt dx x(t + Δt) − x(t) = lim = lim(2At + AΔt) = 2At dt Δt→0 Δt Δt→0 Generalization for Polynomials: x(t) = At n dx = nAt n−1 dt Concept Question: Instantaneous 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 tB, both trains have the same velocity. 2. Both trains speed up all the time. 3. Both trains have the same velocity at some time before tB, . 4. Somewhere on the graph, both trains have the same acceleration. Average Acceleration Change in instantaneous velocity divided by Δt = t − t the time interval 2 1 Δv Δv (vx,2 − vx,1) Δv a ≡ = x ˆi = ˆi = x ˆi = a ˆi ave t t t t ave,x Δ Δ Δ Δ The x-component of the average acceleration Δvx aave,x = Δt Instantaneous Acceleration and Differentiation For each time interval Δ t , calculate the x-component of the average acceleration a (t) = Δv / Δt ave,x x Take limit as Δ t → 0 sequence of the x-component average accelerations Δvx vx (t + Δt) − vx (t) dvx lim = lim ≡ Δt→0 Δt Δt→0 Δt dt The limiting value of this sequence is x-component of the instantaneous acceleration at the time t. a (t) = dv / dt x x Instantaneous Acceleration The x-component of acceleration is equal to the slope of the tangent line of the graph of the x-component of the velocity vs. time at time t dvx ax (t) = dt Group Problem: Model Rocket A person launches a home-built model rocket straight up into the air at y = 0 from rest at time t = 0 . (The positive y- direction is upwards). The fuel burns out at t = t0. The position of the rocket is given by ⎧1 a 2 0 6 4 y = ⎨ (a0 − g)t − t / t0 ; 0 < t < t0 2 30 ⎩ with a0 and g are positive. Find the y-components of the velocity and acceleration of the rocket as a function of time. Graph ay vs t for 0 < t < t0. Non-Constant Acceleration and Integration Change in Velocity: Integral of Acceleration Consider some time t such that t < t < t 0 f Then the change in the x-component of the velocity is the integral of the x- component acceleration (denote v ≡ v ( t ) ). x,0 x 0 t′=t v (t) − v = a (t′)dt′ x x,0 ∫ x t t ′= 0 “Integration is the inverse operation of differentiation” Change in Position: Integral of Velocity Area under the graph of x-component of the velocity vs. time is the displacement (denote x ≡ x ( t )). 0 0 t′=t x(t) − x = v (t′)dt′ 0 ∫ x t t ′= 0 Worked Example: Time-Dependent Acceleration a (t) At 2 Acceleration is a non-constant function of time x = with t = 0 , v = 0 , and x = 0 . 0 x,0 0 Change in velocity: t t t′=t ′= t′3 At3 v (t) − 0 = At′2 dt′ = A ⇒ v (t) = x ∫ 3 x 3 t′=0 t′=0 Change in position: t′=t t t ′= ⎛ t′3 ⎞ ⎛ ⎛ t′4 ⎞ ⎞ At 4 x(t) − 0 = ∫ A⎜ ⎟ dt′ ⇒ x(t) = ⎜ A⎜ ⎟ ⎟ = t 0 ⎝ 3 ⎠ ⎝ ⎝ 12 ⎠ ⎠ 12 ′= 0 Generalization for Polynomials: t′=t t′=t t′n+1 t n+1 t n+1 t′n dt′ = = − 0 ∫ n +1 n +1 n +1 t′=t0 t′=t0 Special Case: Constant Acceleration Acceleration: ax = constant t′=t Velocity: v (t) − v = a dt′ = a t ⇒ x x,0 ∫ x x t′=0 v (t) = v + a t x x,0 x Position: t′=t x(t) − x = (v + a t′)dt′ ⇒ 0 ∫ x,0 x t′=t0 1 2 x(t) = x0 + vx,0t + axt 2 Concept Question: Integration A particle, starting at rest at t = 0, experiences a non- constant acceleration ax(t) . It’s change of position can be found by 1. Differentiating ax(t) twice. 2. Integrating ax(t) twice. 2 3. (1/2) ax(t) times t . 4. None of the above. 5. Two of the above. Group Problem: Sports Car At t = 0 , a sports car starting at rest at x = 0 accelerates with an x-component of acceleration given by 3 1/2 ax (t) = At − Bt , for 0 < t < (A / B) and zero afterwards with A, B > 0 (1) Find expressions for the velocity and position vectors of the sports car as functions of time for t >0. (2) Sketch graphs of the x-component of the position, velocity and acceleration of the sports car as a function of time for t >0 Appendix: Integration and the Riemann Sum Change in Velocity: Area Under Curve of Acceleration vs. time Mean Value Theorem: For each rectangle there exists a time t < t = c < t i i i+1 such that dvx vx (ti+1) − vx (ti ) = (ci )Δt = ax (ci )Δt dt Apply Mean Value Theorem vx (t1) − vx (t0 ) = ax (c0 )Δt (vx (t2 ) − vx (t1)) = ax (c1)Δt ⋅⋅⋅ = ⋅⋅⋅ vx (ti+1) − vx (ti ) = ax (ci )Δt ⋅⋅⋅ = ⋅⋅⋅ v (t ) − v (t ) = a (c )Δt x n x n−1 x n−1 We can add up the area of the rectangles and find i=n−1 v (t ) v (t ) ((a (c ) t) x n − x 0 = ∑ x i Δ i=0 Change in Velocity: Integral of Acceleration The area under the graph of the x-component of the acceleration vs. time is the change in velocity i=N vx (t f ) − vx (t0 ) = lim ax (ti )Δti Δt →0 ∑ i i=1 t=t f v (t ) − v (t ) ≡ a (t)dt x f x 0 ∫ x t t = 0 http://giphy.com/gifs/math-mathematics-calculus-zTGUlIASZx83u .