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
and y = (4.00 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 vectorsiˆ and. By taking derivatives, obtain expressions for (b) the velocity vector as a function of time and (c) the acceleration vector a as a function of time. Next use unit-vector notation to write expressions for (d) the position, (e) the velocity, and (f) the acceleration of the golf ball, all at t = 3.00 s.

Explanation / Answer

x=18t

y=4t-4.9t2

a) position vector of ball=P= xi+yj

=18ti+(4t-4.9t2)j

b) velocity vecor as a function of time= V= x'i+y'j

where x', y' are the derivatives of x, y w.r.t. time t, i.e.

x'=18;y'=4-9.8t

V=18i+(4-9.8t)j

c)accelaration vector a=x''i+y''j

where x''=0; y''=-9.8

a=-9.8j

d) when t=3s, from the above expressions

P(3)=54i-32.1j

e)V(3)=18i-25.4j

f)a(3)=-9.8j

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