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 = 17.7t
and
y = 4.28t ? 4.53t2,
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 ihatbold and jhatbold. (Give the answer in terms of t.)
rarrowbold =

By taking derivatives, do the following. (Give the answers in terms of t.)
(b) obtain the expression for the velocity vector varrowbold as a function of time
varrowbold =

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

(d) Next use unit-vector notation to write expressions for the position, the velocity, and the acceleration of the golf ball at t = 2.79 s.
rarrowbold = m
varrowbold = m/s
aarrowbold = m/s2

Explanation / Answer

r = 17.7t i+ (4.28t -4.53t^2)j m v = 17.7 i + (4.28 -9.06t)j m/s a= 9.06 j m/s^2 at t = 2.79s r = 49.383 i - 23.57 j m v = 17.7 i -20.453j m/s a= 9.06 j m/s^2 |r| = 54.71 |v| = 27.04 |a|= 9.06 hence in terms of unit vector .... where unit vector is given by ====> unit vector = vector / magnitude

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