UNIVERSITY PHYSICS I CHAPTER 3 Kinematics i Rectilinear motion Chapter 3 Kinematics i: rectilinear motion Motion implies change, and it is change make life-and physics-visible and interesting. Kinematics is the theory about the description of motion Physical theories are creations of the human intellect; they must be invented rather than discovered
1 Chapter 3 Kinematics I: rectilinear motion Kinematics is the theory about the description of motion. Physical theories are creations of the human intellect; they must be invented rather than discovered. Motion implies change, and it is change make life-and physics-visible and interesting
83.1 Position and displacement of a particle 1. Several concepts Frame of reference -any object that is chosen for reference of a motion Coordinate system-the abstract of reference Particle-a single mass point without shape. + An object whose part are all move in exactly the same way can be treated as a particle; or a complex object can be treated as a particle if there are no internal motions or the internal motions can be neglected for the problem which you are discussing 83.1 Position and displacement of a particle 2. The position vector and displacement vector of a particle in rectilinear motion Hi=il Displacement--the change of the position vector A=r-1=(x s-Ci)i=4xi Note: The magnitude of the displacement is not necessarily equal to the total distance as traveled by the particle during the time interval At. when Ar=0. As may be not zero
2 §3.1 Position and displacement of a particle 1. Several concepts Frame of reference —any object that is chosen for reference of a motion. Particle—a single mass point without shape. An object whose part are all move in exactly the same way can be treated as a particle; or a complex object can be treated as a particle if there are no internal motions or the internal motions can be neglected for the problem which you are discussing. Coordinate system—the abstract of reference 2. The position vector and displacement vector of a particle in rectilinear motion x O r x i i i ˆ = r x O r x i f f ˆ = r Note: r r r x x i xi f i f i ˆ ˆ ∆ = − = ( − ) = ∆ r r r Displacement—the change of the position vector §3.1 Position and displacement of a particle 1 The magnitude of the displacement is not necessarily equal to the total distance traveled by the particle during the time interval ∆t .when ∆r = 0 , ∆s may be not zero. r ∆s
83.1 Position and displacement of a particle Note Ar is independent of thethe specifi coordinate system we choose 3 The position of a moving particle is a function of time r(t)=x(t)i □■■■ ④ the path of the particle is a straight line t A graph ofx versus t 83.2 The speed and velocity of a moving particle 1. Average speed and average velocity Ar Define: aveRt ave t 个:d≠EF v≠v 2. Instantaneous velocity H and instantaneous speed (t)=x(t) x(2 x(o r(t+A)=x(t+∠)i
3 Note: 3 The position of a moving particle is a function of time, r r 2 ∆ is independent of the the specific coordinate system we choose. r t x t i ˆ ( ) = ( ) r x t A graph of x versus t §3.1 Position and displacement of a particle 4the path of the particle is a straight line. 2. Instantaneous velocity and instantaneous speed §3.2 The speed and velocity of a moving particle 1. Average speed and average velocity Define: t s v ∆ ∆ ave = t r v ∆ ∆ r r ave = ave ave s r v v r r Q∆ ≠ ∆ ∴ ≠ t x 1 ∆t 2 ∆t x(t) ( )1 x t ( ) 2 x t r r t x t i i ˆ = ( ) = ( ) r r r r t t x t t i f ˆ = ( +∆ ) = ( +∆ ) r r
83.2 The speed and velocity of a moving particle Ar=r(t+At)r(t=x(t+At)-x(tli [x(t+)-x() ∠Lt Instantaneous velocity r(t+4)-r(t) dr v(t)=lim vave lim 一=—l dr→0s r→0s ∠t he component of the velocity: D (t=c T dt 83.2 The speed and velocity of a moving particle The direction of the velocity dx(t) dx(t) 0 same as dr <o opposite of Instantaneous speed v(t)=lim v lim As ds(t) dr→0s ave 4→0stdt ∵limr|=dr=d d→0s Ar dr d 40e/b lim lim ave r→0s At dtdt
4 r r t t r t x t t x t i ˆ ∆ = ( + ∆ ) − ( ) = [ ( + ∆ ) − ( )] r r r t x t t x t i t r v ∆ ∆ ∆ ∆ ˆ [ ( ) ( )] ave + − = = r r i t x t r t r t t r t v t v t t ˆ d d d ( ) ( ) d ( ) lim lim 0s ave 0s = = + − = = → → r r r r r ∆ ∆ ∆ ∆ Instantaneous velocity: The component of the velocity: t x t v t x d d ( ) ( ) = §3.2 The speed and velocity of a moving particle v t s t r t r v r r s t t t ∴ = = = = = = → → → d d d d lim lim lim d d 0s ave 0s 0s r r r r r Q ∆ ∆ ∆ ∆ ∆ ∆ Instantaneous speed: t s t t s v t v t t d d ( ) ( ) lim lim 0s ave 0s = = = → → ∆ ∆ ∆ ∆ The direction of the velocity: 0 d d ( ) > t x t 0 d d ( ) < t x t same as , i ˆ i opposite of ˆ §3.2 The speed and velocity of a moving particle
83.3 The acceleration in rectilinear motion 1.Average acceleration Define:ave== ave rave ∠t 2. Instantaneous acceleration if v;=v,(ti,v=v,(t+4t) Av=v-v,=lv(t+45)-v(x)i 83.3 The acceleration in rectilinear motion then as lim 4v= lim /(t+4t)-v2()i_dv() 4→0sAt→+0s dt or(= dv(t) dv(t): dr(t) d'x(t dt The instantaneous acceleration of a particle is the time rate of change of the velocity vector or the first derivative of the instantaneous velocity vector with respect to time; or the second derivative of the instantaneous position vector with respect to time
5 §3.3 The acceleration in rectilinear motion Define: t v v t v a f i ∆ ∆ ∆ r r r r − ave = = 1. Average acceleration i t v t v t a a i x f x i x ˆ ( ) ( ) ˆ ave ave ∆ − = = r 2. Instantaneous acceleration if v v v v t t v x i v v t i v v t t i f i x x i x f x ˆ [ ( ) ( )] ˆ , ( ) ˆ ( ) = − = + − = = + ∆ ∆ ∆ r r r r r then t v t t v t t v t i t v a x x t t d d ( ) ˆ [ ( ) ( )] lim lim 0s 0s r r r = + − = = → → ∆ ∆ ∆ ∆ ∆ ∆ or i t x t t r t i t v t t v t a x ˆ d d ( ) d d ( ) d d ( ) d d ( ) 2 2 2 2 = = = = r r r r The instantaneous acceleration of a particle is the time rate of change of the velocity vector or the first derivative of the instantaneous velocity vector with respect to time;or the second derivative of the instantaneous position vector with respect to time. §3.3 The acceleration in rectilinear motion
83.3 The acceleration in rectilinear motion The component of the acceleration: a2() dv (t) d'x(t) 3. How to detect and measure the instantaneous acceleration Accelerometer-a plum bob Direction of deviation of plum bob from vertical is opposite to the direction of the acceleration of the particle in the horizontal plane. 83.4 Rectilinear motion with a constant acceleration 83.4 Rectilinear motion with a constant acceleration 1. Some rules ignore the effects of air resistance @the origin of the coordinate could be chosen discretionarily Owrite the constant acceleration as a=a i @choose the initial time instant to be t,=0 ∠t Dlet x(t=x(0)=xo v (ti)=v(0)=vro
6 2 2 d d ( ) d d ( ) ( ) t x t t v t a t x x = = The component of the acceleration: §3.3 The acceleration in rectilinear motion 3. How to detect and measure the instantaneous acceleration Accelerometer—a plum bob Direction of deviation of plum bob from vertical is opposite to the direction of the acceleration of the particle in the horizontal plane. a r θ a r θ §3.4 Rectilinear motion with a constant acceleration 1. Some rules §3.4 Rectilinear motion with a constant acceleration 1ignore the effects of air resistance 2the origin of the coordinate could be chosen discretionarily 3write the constant acceleration as 4choose the initial time instant to be a a ix ˆ = r ti = 0 t t t t ∆ = f − i = 5let 0 0 ( ) (0) ( ) (0) i x i x x x t = x = x v t = v = v
83.4 Rectilinear motion with a constant acceleration 2. Rectilinear motion with a constant acceleration dy fr rom dt We have vx(r) dv (t) (t=v+a. dx(t) Likewise. from =v(t) dt We have dx(t) dt=L(vro +a t)dt x(o=xo+vrot+a,t 2 83.4 Rectilinear motion with a constant acceleration One-dimensional motion with constant acceleration a = constant where a=a v(t)=vro+a,t where v(t)=v,(t)i x(t)=xo+vrot+=,. where r(t=x(t)i Eliminate t in equations about v and x Vx m.(x一
7 2. Rectilinear motion with a constant acceleration from x x a t v t = d d ( ) We have v t v a t v t a t x x x x v t v x x x = + = ∫ ∫ 0 t 0 ( ) ( ) d ( ) d 0 Likewise, from ( ) d d ( ) v t t x t = x We have 2 0 0 0 0 t 0 ( ) 2 1 ( ) d ( ) d ( )d 0 x t x v t a t x t v t v a t t x x t x x x x t x = + + = = + ∫ ∫ ∫ §3.4 Rectilinear motion with a constant acceleration One –dimensional motion with constant acceleration: v t v a t v t v t i x x x x ˆ ( ) where ( ) ( ) = 0 + = r x t x v t a t r t x t i x x ˆ where ( ) ( ) 2 1 ( ) 2 = 0 + 0 + = r a a a i x x where ˆ = constant =r §3.4 Rectilinear motion with a constant acceleration Eliminate t in equations about vx and x 2 ( ) 0 2 0 2 v x − v x = ax x − x
83.5 Geometric interpretations I. The change in the position vector component d dr=v,(t)dt Ar=l dx= v(t)dt () Accelerated motion Constant speed motion 2. The change in the velocity component dv dy =a dt a, (tdt dt (t 83.5 Geometric interpretations a a Accelerated motion Constant acceleration motion 3. What does the negative areas mean? a Accelerated motion
8 §3.5 Geometric interpretations 1. The change in the position vector component ∫ ∫ = ∆ = = f i f i t t x x t x t x dx v (t)dt x dx v (t)dt ( ) ( ) 2. The change in the velocity component ∫ ∫ = ∆ = = f i x f x i t t x v t v t x x x x dv a dt v dv a (t)dt ( ) ( ) t t v (t) x v (t) x i t i f t t f t Accelerated motion Constant speed motion t x v x d d = t v a x x d d = t t a (t) x a (t) x i t i f t t f t Accelerated motion Constant acceleration motion t t a (t) x v (t) x Accelerated motion §3.5 Geometric interpretations 3. What does the negative areas mean?
83.5 Geometric interpretations Rostino 83.5 Geometric interpretations A rebounding ball bearing
9 x a x v x §3.5 Geometric interpretations §3.5 Geometric interpretations A rebounding ball bearing
Exercise 1 a geologist at the top of a 100-meter-deep crevasse cannot resist the temptation to hurl a gneiss rock down to the bottom The rock has an initial downward speed of 100m/s as it leaves the geologists hand when t=0s. Find a. The time interval during which the particle is in flight called the time of flight: and b The speed of the particle at the instant just before impact R.P96 Exercise 2 A rocket is launched from rest from an underwater base a distance of 125m below the surface of a body of water It moves vertically upward with an unknown but assumed constant acceleration. and it reaches the surface in a time of 2.15s when it breaks the surface its engines automatically shut off and it continues to rise What maximum height does it reach? (H.P29)
10 Exercise 1 A geologist at the top of a 100-meter-deep crevasse cannot resist the temptation to hurl a gneiss rock down to the bottom. The rock has an initial downward speed of 10.0m/s as it leaves the geologist’s hand when t =0s. Find a. The time interval during which the particle is in flight, called the time of flight; and b. The speed of the particle at the instant just before impact. R. P96 A rocket is launched from rest from an underwater base a distance of 125m below the surface of a body of water. It moves vertically upward with an unknown but assumed constant acceleration , and it reaches the surface in a time of 2.15s. When it breaks the surface its engines automatically shut off and it continues to rise. What maximum height does it reach? Exercise 2 (H. P29)