a watermelon seed has the following coordinates: x= -5.0, y = 8.0m, and z = 0m. find its position vector (a) in unit-vector notation and as (b) a magnitude and (c) an angle relative to the positive direction of the x axis (d) sketch the vector on the right-handed coordinate (3.00m,0m,0m),what is its displacement (e) in unit-vector notation and as (f) a magnitude and (g) an angle relative to the positive x direction?
Chapter 4
4.1 A watermelon seed has the following coordinates: x = -5.0m, y = 8.0m, and z = 0m. Find its position vector (a) in unit-vector notation and as (b) a magnitude and (c) an angle relative to the positive direction of the x axis. (d) Sketch the vector on a right handed coordinate system. If the see is moved to the xyz coordinates (3.00m, 0.0m, 0.0m), what is its displacement (e) in unit-vector notation and as (f) a magnitude and (g) an angle relative to positive direction of the x axis.
(a) In unit vector notation
(b), (c), and (d). Its useful to now to draw a picture
(e) We write the final position and then use it to compute the displacement
(f) and (g). We can now write the displacement as a magnitude and direction
4.6 An ion’s initial and final position vectors are:
The change in position occurs over 10 s. What is the average velocity?
4.7 The position of an electron is given by
(a) What is the electron’s velocity v(t)? At t = 2.00s what is v in (b) in unit-vector notation and as (c) a magnitude and (d) an angle relative to the positive direction of the x axis.
(a) We find the velocity.
(b) At t = 2s.
4.9 A particle moves so that its position (in meters) as a function of time in seconds is
Write expressions for (a) its velocity and (b) its accelerations as a function of time.
(a) Velocity is the derivative of the position
(b) The acceleration is the derivative of the velocity vector.
4.18 A small boll rolls horizontally off the edge of a tabletop that is 1.20 m high. It strikes the floor at a point 1.52 m horizontally away from the edge of the table. (a) How long is the ball in the air? (b)What is the spead at the instant it leaves the table.
a) We begin by writing the conditions that are given.
We now compute how long it was in the air
b) Now that we know the time, we can find the horizontal velocity
4.19 A baseball leaves a pitcher’s hand horizontally at a speed of 161km/hr. The distance to the batter is 18.3 m. (Ignore the effect of air resistance). (a) How long does the ball take to travel the first half of that distance. (b) The second half? (c ) How far does the ball fall in ther first half? (d) During the second half? (e) Why aren’t the quantities in (c) and (d) equal?
a) The ball travels with constant x velocity since there is no acceleration in the horizontal direction if we ignore friction. For the first half (9.15m)
b) Since the ball travels with constant velocity in the x direction, it takes the same time to travel the second half of the trip.
c) During the first half of the trip
d) We compute the total fall for the trip and then subtract the fall during the first half.
e) The distances are different because the ball is accelerating downward.
4.25. A rifle that shoots bullets at 460m/s is to be aimed at a target 45.7 away and level with the rifle. How high above the target must the rifle barrel be pointed so that the bullet hits the target?
We wish to find the distance h. If we can find the angle that the rifle has, we will have found h.
We first solve for the time of flight, and then use that time to find the angle.
Now that we know the angle, we can find h
4.36 A soccer ball is kicked from the ground with an inital speed of 19.5 m/s at an upward angle of 45 degrees. A player 55m away in the direction of the kick starts running to meet the ball at that instant. What must be his average speed if he is to meet the ball just before it hits the ground.
We begin by writing down the things we know.
We solve for the time of flight.
Now we can use the time of flight to find the landing point
The player must run from 55m to 38.75m in 2.81s. His average speed must be
4.43 An earth satellite moves in a circular orbit 640km above Earth’s surface with a period of 98min. What are (a) the speed and (b) the magnitude of the centripetal acceleration of the satellite?
4.46 A carnival merry-go-round rotates abouta vertical axis at a constatn rate. A passenger standing onthe edge of the merry-go-round has a constant speed of 3.66m/s For each of the following instantaneous situations, state how far the passenger is from the cnter of the merry-go-round and in which direction. (a) The passenger has an acceleration of 1.83 m/s2 east. (b) The passenger has an acceleration of 1.83 m/s2 south.
a) We can compute the distance from the magnitude of the acceleration
For the acceleration to be inward and to the east, the passenger must be on the west side.
b) The magnitude of the acceleration is the same, so the passenger remains at 7.32m from the center. For the acceleration to be south, the passenger must be on the north side.
4.53 A cameraman on a pickup truck is travelling westward at 20km/h while he videotapes a cheetah that is moving westward at 30km/h fater than the truck. Suddenly,, the cheetah stops,turns and then runs at 45 km/h eastward as measured by a suddenly nervous crew member who stands alongside the cheetah’s path. The change in the animals velocity takes 2s. What is the acceleration from the perspective of (a) the cameraman and (b) the nervous crew member.
a) Initially the cameraman sees the cheetah going 20 km/h away from him. After the turnaround, the cheetah is going 45 km/h to the east relative to the ground. The cameraman, going west, now sees the cheetah coming toward him at 45+20 =65km/h. The change in velocity (from 20km/h away to 65km/h toward) is 85km/h, or 23.61m/s. The cheetah’s acceleration is 23.61m/s / 2s=11.8m/s2.
b) The crew member sees the same acceleration. The acceleration observed is the same in all inertial (non accelerating) reference frames.
4.54 A boat is traveling upstream at 14km/h with respect to the water in the river. The water is flowing at 9km/h with respect to the ground. (a) What is the velocity of the boat with respect to the ground. (b) A child on the boat walks from front to back at 6km/h with respect to the boat. What is the child’s velocity with respect to the ground.
a) The velocity of the boat with respect to the ground is 14km/h - 9 km/h=5km/h upstream.
b) The velocity of the child with respect to ground is 5km/h -6km/h =-1 km/h (1km/h downstream).
(a) The position vector in unit-vector notation can be written as:
r = -5.0i + 8.0j + 0k
(b) The magnitude of the position vector is given by:
|r| = √((-5.0)^2 + (8.0)^2 + 0^2) = √(25 + 64) = √89
(c) The angle relative to the positive direction of the x-axis can be found using:
θ = arctan(y/x)
θ = arctan(8.0 / -5.0)
(d) To sketch the vector, we start from the origin (0, 0, 0) and move -5.0 units in the x-direction, 8.0 units in the y-direction, and 0 units in the z-direction.
(e) The displacement vector from (3.00m, 0m, 0m) to (-5.0m, 8.0m, 0m) can be found by subtracting the initial position vector from the final position vector:
Δr = (-5.0 - 3.00)i + (8.0 - 0)j + (0 - 0)k
Δr = -8.0i + 8.0j + 0k
(f) The magnitude of the displacement vector is given by:
|Δr| = √((-8.0)^2 + (8.0)^2 + 0^2) = √(64 + 64) = √128 = 8√2
(g) The angle relative to the positive x-direction can be found using the displacement vector:
θ = arctan((8.0 - 0) / (-8.0 - 3.00)) = arctan(8.0 / -11.0)
To find the position vector of the watermelon seed with the given coordinates, follow these steps:
Step 1: Write the coordinates in unit-vector notation:
(a) In unit-vector notation, the position vector is given as:
r = x * i + y * j + z * k
where i, j, and k are the unit vectors along the x, y, and z axes, respectively.
Given: x = -5.0, y = 8.0m, and z = 0m
So, the position vector in unit-vector notation is:
r = -5.0i + 8.0j + 0k
Step 2: Calculate the magnitude of the position vector:
(b) The magnitude of a vector is given by the formula:
|v| = sqrt(vx^2 + vy^2 + vz^2)
where vx, vy, and vz are the components along the x, y, and z axes, respectively.
For the given position vector, the magnitude is:
|r| = sqrt((-5.0)^2 + (8.0)^2 + 0^2)
|r| = sqrt(25 + 64 + 0)
|r| = sqrt(89)
Step 3: Find the angle relative to the positive x-axis:
(c) The angle can be determined by using the formula:
θ = tan^(-1)(vy / vx)
where vy and vx are the components of the vector along the y and x axes, respectively.
For the given position vector, the angle is:
θ = tan^(-1)(8.0 / -5.0)
Step 4: Sketch the vector on the right-handed coordinate system:
(d) On the right-handed coordinate system, the x-axis is positive to the right, the y-axis is positive upwards, and the z-axis is positive coming out of the plane.
To sketch the vector (-5.0, 8.0, 0) on the coordinate system, start at the origin and move 5 units to the left along the x-axis and then 8 units upwards along the y-axis. The vector will be in the second quadrant of the coordinate system.
Now, let's find the displacement of the vector from the point (3.00m, 0m, 0m) to the watermelon seed.
Step 5: Calculate the displacement vector:
(e) The displacement vector can be found by subtracting the initial position vector from the final position vector.
Given: Initial position vector = (3.00, 0, 0)
Final position vector = (-5.0, 8.0, 0)
The displacement vector in unit-vector notation is:
Δr = Final position vector - Initial position vector
Δr = (-5.0 - 3.00)i + (8.0 - 0)j + (0 - 0)k
Step 6: Calculate the magnitude of the displacement vector:
(f) The magnitude of the displacement vector can be found using the same formula as the magnitude of the position vector.
For the displacement vector, the magnitude is:
|Δr| = sqrt(((-5.0 - 3.00)^2) + ((8.0 - 0)^2) + (0 - 0)^2)
Step 7: Find the angle of displacement relative to the positive x-axis:
(g) The angle can be determined using the same formula as finding the angle of the position vector relative to the x-axis.
For the displacement vector, the angle is:
θ = tan^(-1)((8.0 - 0) / (-5.0 - 3.00))
I hope this explanation helps you understand how to find the position vector, displacement vector, and their properties.