Question

A golf ball is hit off a tee at the edge of a cliff. Its x and y coordinates as functions of time are given by x = 17.5t and y = 3.80t-4.90t. where x and y are in meters and t is in seconds (a) Write a vector expression for the ball's position as a function of time, using the unit vectors i and j. (Give the answer in terms of t.) By taking derivatives, do the following. (Give the answers in terms of t.) (b) obtain the expression for the velocity vector v as a function of time m/s (c) obtain the expression for the acceleration vector a as a function of time m/s2 (d) Next use unit-vector notation to write expressions for the position, the velocity, and the acceleration of the golf ball at t-3.21 s. m/s m/s2 a F

Answer 1

Velocity is given as v = dx/dt, acceleration is given as a = dv/dt
Velocity is given as v = dx/dt, acceleration is given as a = dv/dt x = 17.5 t y = 3.80 t - 4.90 t^2 (a) r^rightarrow = x i^+ y j^r^rightarrow = {17.5 t i^+ (3.80 t - 4.90 t^2) j^} m (b) velocity along x-axis in v_x = dx/dt = 17.5 m/s along y-axis in v_y = dy/dt = d/dt (3.80 t - 4.90 t) = 3.80 - 9.80 t therefore v^rightarrow = v_x i^+ v_y j^v^rightarrow = {17.5 i^+ (3.80 - 9.80 t) j^} m/s (c) Now acceleration along x-axis I s a_x = dv_x/dt = 0 & along Y-axis is a_y = dv_y/dt = 0 - 9.80 = -9.80 m/s^2 therefore a^rightarrow a_x i^+ a_y j^a^rightarrow = -9.80 j^m/s^2 (d) Now t = 3.21 sec r^rightarrow = 17.5 times 3.21 i^+ (3.80 times 3.21 - 4.90 times 3.21)^2 j^r^rightarrow = 56.175 i^- 3.531 j^m \r\n v^rightarrow = 17.5 i^+ (3.80 - 9.80 times 3.21) j^m/s