The Position, Velocity, and Acceleration Vectors Problems and Solutions

 Problem#1

A motorist drives south at 20.0 m/s for 3.00 min, then turns west and travels at 25.0 m/s for 2.00 min, and finally travels northwest at 30.0 m/s for 1.00 min. For this 6.00-min trip, find:

(a) the total vector displacement,

(b) the average speed, and

(c) the average velocity. Let the positive x axis point east.

Answer:
the route of motion of a motorist discussion below

Fig.1

From figure 1:
x = –3000 m – 1800 m x cos 45⁰ = –4273 km
y = –3600 m + 1800 m sin 45 = –2327 km

(a) the total vector displacement is
R = √(x² + y²) = √{(–4273 m)² + (–2327 m)} = 4.87 km

(b) the average speed is

average speed = (3600 m + 3000 m + 1800 m)/(180 s + 120 s + 60.0 s) = 23.3 m/s

(c) the average velocity is

average velocity = 4.87 x 10³ m/360 s = 13.5 m/s along R

Problem#2
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 vectors i and j. By taking derivatives, obtain expressions for

(b) the velocity vector v 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.

Answer:

(a) position vector is written as

r = xi + yj

r = 18.0ti + (4.00t – 4.90t²)j

(b) the velocity vector v as a function of time given by

v = dr/dt

v = 18.0i + (4.00 – 9,80t)j

(c) the acceleration vector a as a function of time given by

a = dv/dt

a = (–9,80 m/s²)j

(d) the position,

r(t = 3.00 s) = (18.0 x 3.00)i + (4.00 x 3.00 – 4.90(3.00)²)j

r(t = 3.00 s) = 54.0i + (–32.1)j

(e) the  velocity

v (t = 3.00 s) = 18.0i + (4.00 – 9,80 x 3.00)j

v (t = 3.00 s) = 18.0i + 25.4j

(f) the acceleration of the golf ball is

a = (–9,80 m/s²)j

Problem#3
When the Sun is directly overhead, a hawk dives toward the ground with a constant velocity of 5.00 m/s at 60.0° below the horizontal. Calculate the speed of her shadow on the level ground.

Answer:
The sun projects onto the ground the x-component of her velocity:

v_x = (5.00 m/s) cos (–60.0⁰) = 2.50 m/s

Problem#4
The coordinates of an object moving in the xy plane vary with time according to the equations

x = –(5.00 m) sin ωt and

y = (4.00 m) – (5.00 m) cos ωt, 

where ω is a constant and t is in seconds.

(a) Determine the components of velocity and components of acceleration at t = 0.

(b) Write expressions for the position vector, the velocity vector, and the acceleration vector at any time t > 0 . (c) Describe the path of the object in an xy plot.

Answer:
(a) from x = –(5.00 m) sin ωt, the x-component of velocity is

v_x = dx/dt = –(5.00ω) cos ωt

and a_x = dv_x/dt = +5.00ω² sin ωt

similarly,
v_y = dy/dt = d{(4.00 m) – (5.00 m) cos ωt }/dt = +5.00ω sin ωt

and a_y = dv_y/dt = 5.00ω² cos ωt

at t = 0,

v = v_xi + v_yj

v = (–5.00ω cos ωt)i + (5.00ω sin ωt)j

v(t = 0) = –5.00ωi

and

a = a_xi + a_yj

a = (5.00ω² sin ωt)i + (5.00ω² cos ωt)j

a(t = 0) = 5.00ω²j

(b) the position vector, the velocity vector, and the acceleration vector at any time t > 0,

r = xi + yj = (4.00 m)j + (5.00 m)(–sin ωt i – cos ωtj)

v = dr/dt = (5.00ω m)(–cos ωt i + sin ωtj)

and
a = (5.00ω² m)(sin ωt i + cos ωtj)