Displacement, Velocity and Acceleration
In classical mechanics, kinematics deals with the motion of a particle. It deals only with the position, velocity, acceleration, and displacement of a particle. It has no concern about the source of motion.
Linear Displacement
The term "displacement" refers to when something shifts away from its original "location," and "linear" refers to a straight line. As a result, “Linear Displacement” can be described as the movement of an object in a straight line along a single axis, for example, from side to side or up and down. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Non-contact sensors such as LVDTs and other linear location sensors can calculate linear displacement. Linear displacement is usually measured in millimeters or inches and may be positive or negative.
![obtain a relation
we can combine Equations 20
result is worth the effort!
We first solve Equation 2.6 for r and then sull
tion 2.10 and simplify:
. (Dr. = Voc) + $as (ºr = var) ²2.
a
ax
We transfer the term to to the left side and multiply through by 2a,:
U₂ - Dar
a.
x = x + Uor
2a, (x-xo) = 2000 - 200² + U² - 2uarvx + vaz.
Finally, we combine terms and simplify to obtain the following equation:
Velocity as a function of position for an object moving with constant
acceleration
For an object moving in a straight line with constant acceleration ax,
v² = vo?² + 2ax(x - xo).
Units: The units of each term in this equation reduce to m²/s².
Notes:
EXAMPLE 2.7 Entering the freeway
Now we will apply Equation 2.11 to a problem in which time is neither given nor asked for. A sports car is
sitting at rest in a freeway entrance ramp. The driver sees a break in the traffic and floors the car's accelera-
tor, so that the car accelerates at a constant 4.9 m/s² as it moves in a straight line onto the freeway. What
distance does the car travel in reaching a freeway speed of 30 m/s?
SOLUTION
SET UP As shown in Figure 2.19, we place the origin at the initial posi-
tion of the car and assume that the car travels in a straight line in the +x
direction. Then Vox = 0, v, 30 m/s, and ax = 4.9 m/s².
. This equation is generally used in problems that neither give a time t nor ask for one.
• Vox is the initial velocity of the object when it is at its initial position xo.
• U, is the final velocity of the object when it is at its final position x.
SOLVE The acceleration is constant; the problem makes no mention
of time, so we can't use Equation 2.6 or Equation 2.10 by itself.
(2.11)
-
v² = voz + 2ax(x − xo).
x-xo.=
the timer first
Rearranging and substituting numerical values, we obtain
v² - voz
(30 m/s)² - 0
2ax
2(4.9 m/s²)
= 92 m.
Equat
Any kine
accelera
Video Tutor Solution
We need a relationship between x and Ux, and this is provided by
Equation 2.11:
Equa
CONTINUED](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F28764828-4d1c-44b8-9544-7d79c8b69e6b%2Fce1bcaa2-547b-4ac1-8c68-3c10466a0847%2Fmxo9205_processed.jpeg&w=3840&q=75)
![multiply
we get
ilt we
-nstant
ravel?
ed to
ship,
at the
qua-
Oxo
Vor=0
1
->
a = 4.9 m/s²
Ux - Vox
t =
ax
x = ?
A FIGURE 2.19 A car accelerating on a ramp and merging onto
a freeway.
30 m/s 0
4.9 m/s²
Alternative Solution: We could have used Equation 2.6 to solve for
the time t first:
30 m/s
= 6.12 s.
Equations of motion for constant acceleration
Ux= Vox + axt
x = xo + voxt + 1/{axt²2
v² = vo²² + 2ax(x − xo)
Then Equation 2.10 gives the distance traveled:
RIPO x = xo = Vot + a,²
2.4 Motion with Constant Acceleration
REFLECT We obtained the same result in one step when we used
Equation 2.11 and in two steps when we used Equations 2.6 and 2.10.
When we use them correctly, Equations 2.6, 2.10, and 2.11 always give
consistent results.
The final speed of 30 m/s is about 67 mi/h, and 92 m is about
01100 yd. Does this distance correspond to your own driving experience?
Practice Problem: What distance has the car traveled when it has
reached a speed of 20 m/s? Answer: 41 m.
madura su
sdt ob
= 0 + (4.9 m/s²) (6.12 s)² = 92 m.
Equations 2.6, 2.10, and 2.11 are the equations of motion with constant acceleration.
Any kinematic problem involving the motion of an object in a straight line with constant
acceleration can be solved by applying these equations. Here's a summary of the equations:
vis
(2.6)
(2.10)
hye (2.11)
xo
43
o
The plot with constant acceleration:
x = xo + vox² + 1a,1²
The effect of
acceleration:
a1²
The plot we would get
with zero acceleration:
x = xo +Vox¹
ZOR](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F28764828-4d1c-44b8-9544-7d79c8b69e6b%2Fce1bcaa2-547b-4ac1-8c68-3c10466a0847%2F8kwvp7_processed.jpeg&w=3840&q=75)
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