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 = (16.0 m/s)t
y = (3.40 m/s)t - (4.90 m/s2)t2
(a) Write a vector expression for the ball's position as a function of time, using the unit vectors i and j. (Use t, i and j as necessary.)
r = m
(b) By taking derivatives, obtain an expression for the velocity vector v as a function of time.
v = m/s
(c) By taking derivatives, obtain an expression for the acceleration vector a as a function of time.
a = m/s2
(d) Next use unit vector notation to write an expression for the position of the golf ball at t = 2.30 s.
r(2.30 s) = ( m ) i + ( m ) j
(e) Write an expression for the velocity at this time.
v(2.30 s) = ( m/s ) i + ( m/s ) j
(f) Write an expression for the acceleration at this time.
a(2.30 s) = ( m/s2 ) j

Explanation / Answer

a)r=(16t)i+(3.40t -4.90t2)j

b)v=dr/dt=16i+(3.4-4.9*2t)j=16i+(3.4-9.8t)j

c)a=dv/dt=-9.8j

d)r at t=2.3s

r=16*2.3i+(3.4*2.3-4.9*2.3*2.3)j=36.8i+18.101j m

e)v at 2.3s

v=16i+(3.4-9.8*2.3)j=16i+19.14j m/s

f)a at t=2.3s

a=-9.8j m/s^2

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