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 = 18.0t and y = 4.00t – 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 and . By taking derivatives, obtain expressions for (b) the velocity vector as a function of time and (c) the acceleration vector as a function of time. (d) Next use unit-vector notation to write expressions for the position, the velocity, and the acceleration of the golf ball at t = 3.00 s.

Answer 1

Position:                 x = 18t    y = 4t - 4.9t²

First derivative:        x' = 18      y' = 4 - 9.8t

Second derivative:    x'' = 0        y'' = - 9.8


Position vector:      P  =  (18t) i  +  (4t - 4.9t²) j

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

Acceleration vector  A  =              (- 9.8) j

Answer 2

Explanation:

It is given that,

A golf ball is hit off a tee at the edge of a cliff. The x component is given by, x = 18 t

y component is given by, y = 4 t - 4.9 t²

Where x and y are in meters and t is in seconds.

(a) The ball’s position as a function of time, using the unit vectors is given by :

$r=18t+(4t-4.9t^2)$

(b) Velocity, $v=\dfrac{dr}{dt}$

$v=\dfrac{d(18t+(4t-4.9t^2)}{dt}$

$v=18+(4-9.8t)$

i.e. v = 18 i +(4 - 9.8t) j

(c) Acceleration vector, $a=\dfrac{dv}{dt}$

$a=\dfrac{d(18i+(4-9.8t)j}{dt}$

a = -9.8 j m/s²

(d) Expressions for the position at t = 3 s

$r=18(3)i+(4(3)-4.9(3)^2)j$

r = (54 i - 32.1 j) m

Expressions for the velocity at t = 3 s

v = 18 i +(4 - 9.8(3)) j

v = (18 i -25.4 j) m/s

Expressions for the acceleration at t = 3 s

a = -9.8 m/s² j

Hence, this is the required solution.