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 byx 19.3t and y 4.16t 4.90t2, 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.) r=| ( 19.31)i-t1.161-1.9012)jX m 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 v=| (19·3 ) i +(4.16-9.80t j |X m/s (c) obtain the expression for the acceleration vector a as a function of time a9.8 (d) Next use unit-vector notation to write expressions for the position, the velocity, and the acceleration of the golf ball at t- 2.97 s r-57.321,-30.867/ X m 57.32 1it33.266j|X m/s a =|-9.8

Explanation / Answer

(a) r(t) = 19.3 m/s ti + (4.16 m/s * t - 4.90 m/s/s*t2) j

(b) v(t) = dr/dt = 19.3m/s i + (4.16 m/s - 9.80 m/s/s t) j

(c) a(t) = dv/dt = 0 i - 9.80 m/s/s j = -9.80 m/s/s . j

(d) r(4.40s) = 57.321 m (i) - 30.867(j)
v(4.40s) = 57.321m/s (i) + (-33.266 m/s)(j)
  a(4.40s) = -9.80 m/s2 (j)

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