A golf ball is hit off a tee at the edge of a cliff. Its x and y coordinates as

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 = 18.0t and y = 4.00t - 4.90t^2, 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, By taking derivatives, obtain expressions for (b) the velocity vector v as a function for time and (c) the acceleration vector a as a function of time. (d) Next use unit-vector notion to write expressions for the position, the velocity, and the acceleration of the golf ball at t = 3.00s.

Explanation / Answer

position of ball in x and y coordinates                 x = 18 t                 y = 4 t - 4.9 t^2   a)   x = 18 t          y = ( 4 t - 4.9 t^2 )   b)   velocity v_x = dx / dt                             = d /dt ( 18 t ) = 18 m/s                      v_y = d y / dt                            = d/ dt ( 4 t - 4.9 t^2 )                            = 4-9.8 t   m/s             c) acceleration              a_x = dv_x / dt = 0               a_y = dv_y / dt   = - 9.8 m/s^2 d) unit vector notation for position , velocity and acceleration   at t = 3 s                 position s =   x i + y j                                 = (18 t) i + ( 4 t - 4.9 t^2 ) j    at t = 3s                                 =   54 i  - 32 .1 j m                 velocity v = v_x i + v_y j                                 = 18 i + (4-9.8 t) j                                        =   18 i - 25.4 j m/s                 acceleration a = - 9.8 j m/s^2            

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