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 the following expressions. x = (18.0 m/s)t y = (3.00 m/s)t - (4.84 m/s^2)t^2 Write a vector expression for the ball's position as a function of time, using the unit vectors i and j. (Use the following as necessary: t.) By taking derivatives, obtain an expression for the velocity vector v as a function of time. By taking derivatives, obtain an expression for the acceleration vector a as a function of time. Next use unit vector notation to write an expression for the position of the golf ball at t = 4.90 s. Write an expression for the velocity at this time. Write an expression for the acceleration at this time.

Explanation / Answer

(a) r(t) = 18 m/s ti + (3.00m/s * t - 4.84 m/s/s*t^2) j

(b) v(t) = dr/dt = 18.0 m/s i + (3.00 m/s - 9.68 m/s/s t) j

(c) a(t) = dv/dt = 0 i - 9.68 m/s/s j = -9.68 m/s/s j

(d) r(t) = 18 m/s ti + (3.00m/s * t - 4.84 m/s/s*t^2) j

r(4.90s) = 18 m/s (4.90s)i + (3.00m/s * (4.90s) - 4.84 m/s/s*(4.90s)^2) j
  
=88.2m (i)+ 101.50 (j)

(f) v(t) = dr/dt = 18.0 m/s i + (3.00 m/s - 9.68 m/s/s t) j

v(4.90s) = 18.0 m/s (i) + (-44.43 m/s)(j)

(g) a(4.40s) = -9.68 m/s/s (j)

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