A small metal ball is kicked from the edge of a cliff. Its x - and y -coordinate

Question

A small metal ball is kicked from the edge of a cliff. Its x- and y-coordinates as functions of time are given by

x = 16.4t

and

y = 3.88t 4.90t2,

where x and y are in meters and t is in seconds.

A small metal ball is kicked from the edge of a cliff. Its x-and y-coordinates as functions of time are given by x-16.4t and y = 3.88t-4.90, where x and y are in meters and t is in seconds. (Do not include units in your answer.) (a) Write a vector expression for the ball's position as a function of time (in m), using the unit vectors i and j. (Give the answer in terms of t.) By taking derivatives, do the following (b) Obtain the expression for the velocity vector v as a function of time (in m/s). (Give the answers in terms of t.) m/s (c) Obtain the expression for the acceleration vector a as a function of time (in m/s) m/s (d) Next, use unit-vector notation to write expressions for the position, the velocity, and the acceleration of the metal ball at t- 3.06 s. (Assume the position is in m, the velocity is in m/s and the acceleration is in m/s-.) m/s m/s2

Explanation / Answer

The vector expression for the ball's positon as a function of time is
r(t) = 16.4t i^ + (3.88t 4.90t2) j^
b) The expression for the velocity of the ball as a function of time is the time deriative of the expression for the position
v(t) = 16.4 i^ m/s + (3.88 - 9.8 t ) j^ m/s
a) The expression for the acceleration of the object is the time derivative of the expression for the velocity.
a(t) = - 9.8 j^ m/s2.
d) The position of the ball at the time t = 3.6 s is
r(t) = 16.4t i^ + (3.88t 4.90t2) j^
r (3.6s) = 16.4 x 3.6 i^ + (3.88 x 3.6 4.90 x (3.6)2) j^
r (3.6 s) = 59.04  i^ -  49.536 j^
The expression for velocity
v(t) = 16.4 i^ m/s + (3.88 - 9.8 t ) j^ m/s
v(3.6) = ( 16.4 i^ - 31.4  j^ ) m/s.
The acceleration is constant throughout the motion and is equal to  
a = - 9.8 j^ m/s2

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