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 = 17.9t and y = 3.92t - 5.13t2,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 and . (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 as a function of time
=

(c) obtain the expression for the acceleration vector as a function of time
=

(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
= m/s
= m/s2

Explanation / Answer

   (a)

             Given equations:

                          x = 17.9t and y = 3.92t - 5.13t^2

       Velocity along horizontial direction =   Vx = 17.9(i^)          

       Velocity along the veritical direction = Vy = 3.92 (J^)   

where   i^ , j^ are the unit vectors along the horizontial , vertical directions respectvely

(b)

                   x = 17.9 t   velocity   = Vx = dx / dt = 17.9 (i^)          

                    y = y = 3.92t - 5.13t^2 = Vy = dy/dt = 3.92 - 5.13(2t)

                                                                         =3.92 (J^) - 10.26 t (J^)

(c)

     accleration   = a_x = dVx/dt = 0 (i^) m/s^2

                           a_y = dVy/dt = - 5.13(2) = - 10.26 m/s^2 (J^)  

(d)

   s vlaue not clear..      

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