Agolf 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.

A) The vector expression for the ball’s position as a function of time, using the given unit vectors is; R(t) = (18t)i + (4t - 4.9t²)j

B) The velocity expression for the ball as a function of time, using the given unit vectors is; V(t) = 18i + (4 - 9.8t)j

C) The acceleration expression for the ball as a function of time, using the given unit vectors is; a(t) = -9.8j

D) The position, velocity and acceleration of the ball at t = 3 s are;

R(3) = (54i - 32.1j) m

R(3) = (54i - 32.1j) mV(3) = (18i - 25.4j) m/s

R(3) = (54i - 32.1j) mV(3) = (18i - 25.4j) m/sa(3) = -9.8j m/s²

We are given the x and y coordinates distance in terms of t as;

x(t) = 18t

x(t) = 18ty(t) = 4t - 4.9t²

A) We want to write a vector for the balls' position as a function of time. This simply means the resultant of the unit vectors.

Thus;

R(t) = x(t)i + y(t)j

R(t) = (18t)i + (4t - 4.9t²)j

B) The velocity will be the derivative of the position with respect to time. Thus;

V(t) = R'(t)

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

C) The acceleration vector as a unit of time will be the derivative of the velocity. Thus;

a(t) = V'(t)

a(t) = -9.8j

D) At t = 3 s

R(3) = (18 × 3)i + ((4 × 3) - (4.9 × 3²))j

R(3) = (54i - 32.1j) m

V(3) = 18i + (4 - (9.8 × 3))j

V(3) = (18i - 25.4j) m/s

a(3) = -9.8j

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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 :

(b) Velocity,

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

(c) Acceleration vector,

a = -9.8 j m/s²

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

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.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s