C-sol - Chapter C - Problems Blinn College - Physics 2425 -...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter C - Problems Blinn College - Physics 2425 - Terry Honan Problem C.1 Two points have cartesian coordinates H 3 m, - 8 m L and H - 5 m, 5 m L . (a) Convert these points to polar coordinates. (b) What is the distance between these points? Solution to C.1 (a) To convert from cartesian coordinates to polar use: r = x 2 + y 2 q = : tan - 1 y x when x > 0 180 ° + tan - 1 K y x O when x < 0 H x 1 , y 1 L = H 3, - 8 L ï r 1 = 3 2 + 8 2 = 8.54 m and q 1 = tan - 1 K - 8 3 O = - 69.4 ° H x 2 , y 2 L = H - 5, 5 L ï r 2 = 5 2 + 5 2 = 7.07 m and q 2 = 180 ° + tan - 1 K 3 - 3 O = 135 ° (b) The distance between two points is: H D x L 2 + H D y L 2 = H x 2 - x 1 L 2 + H y 2 - y 1 L 2 = 233 = 15.3 m. Problem C.2 Two points have polar coordinates H r , q L = H 5 m, 25 ° L and H r , q L = H 8 m, 120 ° L . (a) Convert these points to cartesian (rectangular) coordinates. (b) What is the distance between these points? Solution to C.2 (a) To convert from polar to cartesian coordinates use: x = r cos q and y = r sin q H r 1 , q 1 L = H 5 m, 25 ° L ï H x 1 , y 1 L = H 4.53 m, 2.11 m L H r 2 , q 2 L = H 8 m, 120 ° L ï H x 2 , y 2 L = H - 4 m, 6.93 m L (b) The distance between points is
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
D x 2 + D y 2 = H x 2 - x 1 L 2 + H y 2 - y 1 L 2 = 9.77 Problem C.3 A car drives around a traffic circle with a radius of 20 m. (a) When the car drives half way around the circle, what is the magnitude of the displacement vector and the total distance driven? (b) When the car drives once around a full circle, what is the magnitude of the displacement vector and the total distance driven? Solution to C.3 (a) The displacement vector is the vector from the starting position to the stopping point. It's magnitude is then the distance between these two point. In this case the distance is 2 r = 2 μ 20 = 40 m. The total distance driven is half the circumference. 1 2 2 ˛ r = ˛ 20 m = 62.8 m (b) For a full circle the magnitude of the displacement is 0 , so its magnitude is 0. The total distance is the circumference. 2 ˛ r = 2 ˛ 20 m = 126 m Problem C.4 (a) Obtain a component expression for a position vector when its magnitude and direction are given by r = 40 m and q = 155 °. (b) A car moves at 70 mi ê hr in the direction 12° south of west. Obtain a component expression for the velocity of the car. (c) A rope pulls with a force of 40 N in the direction given by q = - 75 °. Obtain a component expression for the force vector. Solution to C.4 The magnitude and direction of a position vector in two dimensions are just its polar coordinates and the component expression for it is r = x x ` + y y ` = X x , y \ = X r cos q , r sin q \ . For any other 2D vector A we can write it in terms of its magnitude A and direction q A = A x x ` + A y y ` = Y A x , A y ] = X A cos q , A sin q \ . (a)
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 14

C-sol - Chapter C - Problems Blinn College - Physics 2425 -...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online