1Non-Uniform Acceleration,Vectors,Kinematics in Two-Dimensions8.01 W02D2Non-Uniform Accelerationthe area under the graph of the x-component of the acceleration vs. time is the change in velocityVelocity as the Integral of the Acceleration010() lim () ( ,)ittiNxxiixtitatdt at t Areaat′==Δ→=′=′′≡Δ=∑∫ax(′t )d′t′t =0′t =t∫=dvxdtd′t′t =0′t =t∫= d′vx′vx=vxt=0()′vx=vxt()∫= vx(t) − vx,0Position as the Integral of VelocityArea under the graph of x-component of the velocity vs. time is the x-component of the displacement()xdxvtdt≡00() ()ttxtvtdt xt x′=′=′′=−∫2Table Problem: Time Dependent AccelerationConsider an object released at time t = 0 with an initial x-component of velocity , located at position ,and accelerating according toFind the velocity and position as a function of time. ax(t) = b0− b1t vx,0 x0Table Problem: Time Dependent AccelerationVelocity:Position: vx(t) = vx,0+ ax(′t )d′t′t =0′t =t∫= vx,0+ (b0− b1′t )d′t′t =0′t =t∫= vx,0+ b0′t′t =0′t =t−12b1′t2′t =0′t =t= vx,0+ b0t −12b1t2 x(t) = x0+ vx(′t )d′t′t =0′t =t∫== x0+ (vx,0+ b0′t −12b1′t2)d′t′t =0′t =t∫= vx,0t +12b0t2−16b1t3Concept QuestionConsider an object released at time t = 0 with an initial x-component of velocity ,and accelerating according toAfter a very long time, the x-component of the velocity is1. Zero2. c0-c13. c0/c14. c0+c15. Not sure vx,0= 0ax≡dvxdt= c0− c1vxVectors3Coordinate System1. An origin as the reference point2. A set of coordinate axes with scales and labels3. Choice of positive direction for each axis 4. Choice of unit vectors at each point in spaceCoordinate system: used to describe the position of a point in space and consists ofCartesian Coordinate SystemVectorA vector is a quantity that has both direction and magnitude. Let a vector be denoted by the symbol The magnitude ofis denoted byAA|A≡A|Vector AdditionABLet and be two vectors. Define a new vector ,the “vector addition” of and by the geometric construction shown in either figure=+CABABSummary: Vector PropertiesAddition of Vectors1. Commutativity2. Associativity3. Identity Element for Vector Addition such that4. Inverse Element for Vector Addition such thatScalar Multiplication of Vectors1. Associative Law for Scalar Multiplication2. Distributive Law for Vector Addition3. Distributive Law for Scalar Addition4. Identity Element for Scalar Multiplication: number 1 such that1=AA()bc b c+=+AAA()ccc+= +AB A B()() ( ) ()bc bc cb cb===AAAA+=+ABBA() ()++=+ +AB CA BC 0+=+ =A00AA −A()+− =AA04Application of Vectors(1) Vectors can exist at any point P in space. (2) Vectors have direction and magnitude.(3) Vector Equality: Any two vectors that have the same direction and magnitude are equal no matter where in space they are located.Vector DecompositionChoose a coordinate system with an origin and axes. We can decompose a vector into component vectors along each coordinate axis, for example along the x,y, and z-axes of a Cartesian coordinate system. A vector at P can be decomposed into the vector sum,xyz=++AA A A Unit Vectors and ComponentsThe idea of multiplication by real numbers allows us to define a set of unit vectors at each point in space withComponents:()ˆˆˆ,,ijk1, 1, 1ˆˆˆ|| || | |===ijkxx yy zzˆˆˆA,A, A== =AiAjAkxyzˆˆˆAAA=++Aijk()xyzA,A ,A=AVector Decomposition in Two DimensionsConsider a vector x- and y components:Magnitude:Direction:(0)xyA,A ,=AAx=Acos(θ), Ay=Asin(θ)A = Ax2+ Ay2AyAx=Asin(θ)Acos(θ)= tan(θ)θ= tan−1(Ay/ Ax)5Vector AdditionVector Sum:Components A = Acos(θA)ˆi + Asin(θA)ˆj B = Bcos(θB)ˆi + Bsin(θB)ˆj=+CAB Cx= Ax+ Bx,Cy= Ay+ ByCx= C cos(θC) = Acos(θA)+ Bcos(θB)Cy= Csin(θC) = Asin(θA) + Bsin(θB)ˆˆ ˆˆ()()cos()sin()xx yy C CAB AB C Cθθ=+ ++ = +Cij ijTable Problem: Displacement VectorAt 2 am one morning a person runs 250 m along the infinite corridor at MIT from Mass Ave to the end of Building 8, turns right at the end of the corridor and runs 178 m to the end of Building 2, and then turns right and runs 30 m down the hall. What is the direction and magnitude of the straight line between start and finish?Table Problem: SolutionTotal displacement:Magnitude:Direction:123ˆˆ ˆ250 m 178 m 30 mˆˆ178 m 220 mΔ=Δ +Δ +Δ=++−=+rrr rjijij 221/2((178 m) (220 m) ) 283 mΔ= + =r11tan (( ) /( ) )tan (220 m /178 m) 51.0yxorrθ−−=ΔΔ==Vector Description of Motion Position Displacement Velocity Accelerationˆˆ() () ()txt yt=+rij() ()ˆˆˆˆ() () ()xydx t dy ttvtvtdt dt=+≡+vijij()()ˆˆˆˆ() () ()yxxydv tdv ttatatdt dt=+≡+aijijˆˆ() () ()txt ytΔ=Δ +Δrij6Constant Acceleration0, 0, ,xxx y yyvvat vvat=+ =+ x = x0+ vx 0t +12axt2, y = y0+ vy 0t +12ayt2 2ax(x − x0) = vx2− vx,02• Components of Velocity: • Components of Position: • Eliminating t: Two-Dimensional MotionProjectile Motion Gravitational Force Law totalintotalintotalinxxyyFmamFma⎧=⎪=⇒⎨=⎪⎩Fagrav grav grav, gravˆymg F mg=− ⇒ =−FjA projectile is fired from a height y0with an initial speed v0at an angle θabove the horizontal. Ignore air resistanceNewton’s Second LawEquations of Motion y-component: x-component: Principle of Equivalence: Components of acceleration:grav inymgma−=in0xma=grav inmm=0,xyaag==−29.8 m/sg =7Concept Q.: 2-dim kinematics A person simultaneously throws two objects in the air. The objects leave the person’s hands at different angles and travel along the parabolic trajectories indicated by A and B in the figure below. Which of the following statements best describes the motion of the two objects? 1. The object moving along the trajectory A hits the ground before the object moving along the trajectory B.2. The object moving along the higher trajectory A hits the ground after the object moving along the lower trajectory B.3. Both objects hit the ground at the same time.4. There is not enough information is specified in order to determine which object hits the ground first.Kinematic Equations --x-components : Acceleration : Velocity : Position :0xa =0,0()xxtxvt=+,0()xxvt v=Kinematic Equations y-components :
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