Physic 231 Lecture 5 Main points of today s lecture g 9 8 m s 2 downwards v v0 gt 1 y y y0 v vo t 2 1 y v0t gt2 2 v2 v02 2g y Here we assumed t0 0 and the y axis to be vertical Two dimension coordinate systems Vectors and components Main points of today s lecture Addition of vectors Zero vectors inverse vectors and vector subtraction Example addition of vectors Trajectories of objects in 2 dimensions g 9 8 m s 2 downwards vy vy0 gt y y y0 1 vy vyo t 2 1 y vy 0t gt2 2 v2y v2y 0 2g y vx vx0 x vx0t More about vectors v A Ax By the Pythagorean theorem the length of this vector is A A A 2 x 2 y The angle can be obtained from various trigonometric relations Ay Ax sin cos A A Ay and tan Ax Ay Vector addition Place the tail of one vector B on the head of the other A Draw the sum vector C from the tail of the first vector to the head of the second v A v B v C v B v A It doesn t matter which order you do it Summation of vectors by components Other vector properties operations Inverse A is drawn in the opposite direction as A The x component of A is Ax and the y component is Ay Obviously R A A A A 0 both from the drawing and also by adding components v A v A Subtracting vectors can be done by subtracting components If R B A then Rx Bx Ax and Ry ByAy Graphically v A v R v A v B Example A football player runs the pattern given in the drawing by the three displacement vectors A B and C The magnitudes of these vectors are A 5 m B 15 m and C 18 m Using the component method find the magnitude and direction of the resultant vector A B C Concept question v If you have a displacement vector A whose v magnitude is a distance of 4 m and add it to another displacement v Bvwithv magnitude 3 m it is possible that the total displacement C A B may have a magnitude of 2 m i e be smaller in magnitude than either A or B a true b false v B v A v v v C A B Actually C can be any number between A B 1m to A B 7m Motion in two dimensions x and y motions must satisfy their own set of equations v y v y 0 a y t v y 0 gt y y y 0 1 v y v yo t 2 1 1 y v y 0t a y t 2 v y 0t gt 2 2 2 v 2y v 2y 0 2a y y 2 g y vx vx 0 axt vx 0 x x x0 1 vx vxo t vx 0t 2 1 x v x 0 t a x t 2 v x 0 t 2 v 2x v x20 2a x x 0 This means that the x component of the velocity remains constant The y component reflects the gravitational acceleration Let s see whether that is true example A bullet is fired from a rifle that is held 1 6 m above the ground in a horizontal position The initial speed of the bullet is 1100 m s Find a the time it takes for the bullet to strike the ground and b the horizontal distance traveled by the bullet If upward is the direction of positive y vx0 1100 m s vy0 0 y 1 6 m x y Concept problem A battleship simultaneously fires two shells at enemy ships If the shells follow the parabolic trajectories shown which ship gets hit first a A b both at the same time c B d need more information initial vertical velocity for A exceeds the initial vertical velocity for B v y2 top v02 2 gh 0 v02 2 gh hA hB v0 y A v0 y B 2v y 0 1 2 ytot 0 v y 0ttot gttot ttot g 2 comparing the two trajectories v0 y A v0 y B ttot A ttot B example A motorcycle daredevil is attempting to jump across as many buses as possible see the drawing The takeoff ramp makes an angle of 18 0 above the horizontal and the landing ramp is identical to the takeoff ramp The buses are parked side by side and each bus is 2 74 m wide The cyclist leaves the ramp with a speed of 33 5 m s What is the maximum number of buses over which the cyclist can jump 33 5 m s 2 74m v0 33 5 m s Which equation for vertical motion 18o should one use a vy vy 0 gt b y y y0 1 vy vyo t 2 1 c y vy 0t gt2 2 d v2y v2y 0 2g y
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