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 = 18.5t
and
y = 3.92t 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 =
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 =
m/s

(c) obtain the expression for the acceleration vector
a
as a function of time
a =
m/s2

(d) Next use unit-vector notation to write expressions for the position, the velocity, and the acceleration of the golf ball at t = 2.73 s.

r
=

m
v
=

m/s
a
=

m/s2

Explanation / Answer

x = 18.5t
y = 3.92t 4.90t^2
(a)
r = xi + yj
r = 18.5t i + (3.92t 4.90t^2) j m
(b)

Velocity v = distance/time
v = dr/dt
v = d/dt(18.5t i + (3.92t 4.90t^2) j) m/s
v = 18.5 i + (3.92 - 9.8t) j m/s

(c)
We know Acceleration a = dv/dt
a = d/dt(18.5 i + (3.92 - 9.8t) j)
a =-9.8 j m/s^2

(d)
At t = 2.73 s
Positon r =  18.5 * 2.73 i + (3.92 *2.73 4.90 * 2.73^2) j m
Positon r = 50.51 i - 25.82 j m

Velocity v = 18.5 i + (3.92 - 9.8*2.73) j m/s
Velocity v = 18.5 i - 22.834 j m/s

Acceleration a = -9.8 j m/s^2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.