Chapter (1)
Rectilinear motion
Introduction
Kinematics of a Particle
The subject of dynamics is classified into the following two branches:
1. Kinematics: In studying the kinematics, we only describe the motion of the
particle without considering the forces that act upon it, i. e, we study the
variations of the particle` s position, velocity, and acceleration with the
time.
2. Kinetics: In kinetics, all the forces that act upon the particle are considered.
We first apply Newton`s second law of motion or the principle of work and
energy to determine the acceleration or the velocity of the particle,
respectively, then, we complete the analysis of the motion by studying the
kinematics.
Rectilinear Motion
In rectilinear motion, the particle moves along a straight line, which is considered
to be the s – axis. The kinematics of the motion is described as follows:
s
O
s
The position ( s ):
The position of the particle is defined by the distance x between the particle and
a fixed origin O on the straight line. The position may be positive (if the particle is
to the right of the origin ) or negative ( if it is to the left ). The position x varies
with the time during the motion, i. e,
s=(t)
where  ( t ) is a certain function of the time t.
The displacement (  s ) :
1
The displacement of a particle during a certain time interval is defined as the
change of its position. If at a certain instant ti the corresponding position is xi
and at another instant tf ( tf  ti ) the corresponding position is sf then, the
displacement that happened during the time interval ( t = tf – ti ) will be
s = sf – si. The displacement x may be positive or negative.
sf
O
ti
si
Δs
𝜟𝒔
The average velocity
The velocity ( v ):
tf
𝒗𝒂𝒗 = 𝜟𝒕
The instantaneous velocity of the particle ( its velocity at any
instant ) is defined by the rate of change of its position with respect to the time
𝜟𝒔 𝒅𝒔
=
𝜟𝒕→𝟎 𝜟𝒕
𝒅𝒕
𝒗 = 𝑳𝒊𝒎
The acceleration ( a ):
It is defined as the time rate of change of the particle`s velocity, i. e,
a=
dv
dt
In some cases, as we will see later, it will be necessary to substitute the
acceleration in the following mathematical form:
𝒅𝒗
𝒂 = 𝒗 𝒅𝒔
Which is derived as follows :
𝒅𝒔
, ∵ 𝒅𝒕 = 𝒗
𝒅𝒗
𝒅𝒗
𝒅𝒔
, 𝒂 = 𝒅𝒕 = 𝒅𝒔 × 𝒅𝒕
𝒅𝒗
∴ 𝒂 = 𝒗 𝒅𝒔
2
s
x
The distance traveled ( D ) :
The distance traveled during a time interval t is defined by the total length of the
path over which the particle travels. To determine the distance traveled during
a certain time interval t from t = ti up to t = tf, the following steps must be
followed:
a- determine the position xi that corresponds to ti and sf that corresponds to
ssf.
b- substitute v = 0 into the relation ( v , t ) to determine the solutions of this
algebraic equation ( v = 0 ) , say t1 , t2,….
c- determine the corresponding positions that correspond to the instants t1, t2
, …, that are contained inside the time interval t.
d- plot, on the straight line, the values of the position s that correspond to the
instants ti , tf, t1, …as shown and calculate the path length between each
two successive instants.
D = d1 + d2 + d3.
+
sf
s1
si
s2
O
t
ti
t1
d1
d2
d3
3
t
s
Definitions:
For a certain time interval between (𝒕 = 𝒕𝟏 , 𝒕 = 𝒕𝟐 ) :
1. Displacement :
∆𝒔|𝒕 →𝒕 = 𝒔𝟐− 𝒔𝟏
𝟏
𝟐
∆𝒔|𝒕 →𝒕
(magnitude and sign).
(magnitude and sign).
𝟏 𝟐
2. Average velocity: 𝒗𝒂𝒗 =
𝒕𝟐 −𝒕𝟏
3. The distance traveled ( D ) :
𝐃 = 𝑻𝒉𝒆 𝒔𝒖𝒎 𝒐𝒇 𝒕𝒓𝒊𝒑 𝒔𝒆𝒈𝒎𝒆𝒏𝒕𝒔 = 𝒅𝟏 + 𝒅𝟐 + 𝒅𝟑 + ⋯ = ∑ 𝒅𝒊
4. Average speed: 𝒗𝒔 =
𝐓𝐡𝐞 𝐝𝐢𝐬𝐭𝐚𝐧𝐜𝐞 𝐭𝐫𝐚𝐯𝐞𝐥𝐞𝐝 ( 𝐃 )
𝒕𝟐 −𝒕𝟏
4
(magnitude only)
Types of applications:
1. Differentiation problems:
Given: The relation between the position and the time ( s , t ).
Required: The velocity v and the acceleration f at any instant.
Method of solution: By differentiating
( s , t ) we obtain (v , t ), then
differentiating ( v , t ), we obtain ( a , t ).
EXAMPLE ( 1 )-Hibbeler R12-1:
A particle is moving along a straight line such that its position is given by :
s = t3 – 6t2 + 9t m, where t is in seconds. Determine the distance traveled ,
displacement , average velocity , and the average speed during the first
2 seconds.
Solution:
s = t3 – 6t2 + 9t
v = ds / dt = 3t2 – 12t + 9
the distance traveled is determined as follows :
s0 = 0
,
s2 = ( 2 )3 – 6 ( 2 )2 + 9( 2 ) = 2 m
when v = 0 :
3t2 – 12t + 9 = 0
t2 – 4t + 3 = 0
(t–1)(t–3)=0
t = 1 sec or
t = 3 sec
the solution t = 3 sec. is outside the time interval from t = 0 to t = 2 sec., then we
determine the position x1 only.
s1 = 1 – 6 + 9 = 4 m
D=4+2=6m,
5
𝜟𝒔
𝜟𝒔 = +𝟐𝒎,
O
+𝟐
𝟏𝒎
𝑫
𝒗𝒂𝒗 = 𝜟𝒕 = 𝟐 = + 𝒔 ,
+
𝟔
𝒗𝒔 = 𝜟𝒕 = 𝟐 = 𝟑𝒎/𝒔
s1 = 4m
t=1
sec.
t = 2 sec.
t=0
s= 0
s2 = 2m
s
4m
𝜟𝒔
2m
EXAMPLE ( 2 ) :
A particle moves along a straight line such that its acceleration is given by:
a = 2t – 6 m / s2, where t is in seconds. If the motion is started from the origin
with a velocity of 5 m / sec., determine the distance it travels during the first
6 seconds.
Solution:
a = dv / dt = 2t – 6
v
t
5
0
dv = ( 2t – 6 ) dt
 dv =  (2 t − 6 )dt
v – 5 = t2 –6t
v = t2 – 6t + 5
ds / dt = t2 – 6t + 5
ds = ( t2 – 6t + 5 ) dt
𝒔
𝒕
∫𝟎 𝒅𝒔 = ∫𝟎 (𝒕𝟐 − 𝟔𝒕 + 𝟓) 𝒅𝒕
t3
s=
3
- 3t
2
+5t
6
To determine the distance traveled during the interval from t = 0 up to t = 6
sec.:
s6 = ( 6 )3/3 - 3( 6 )2 + 5( 6 ) = - 6 m
s0 = 0
when v = 0 :
t2 – 6t + 5 = 0
( t – 1 )( t – 5 ) = 0
t = 1 sec.
s1 = 1/3 –3 + 5 = 7/3 m
t = 5 sec.
or
t = 5 sec.
, s5 = ( 5 )3/3 - 3 ( 5 )2 + 5 ( 5 ) = - 25/3
t=0
s0 = 0 O
t = 6 sec.
s6 = - 6m
s5 = -25/3 m
m
t = 1sec.
s1 = 7/3m
6m
7/3 m
32/3 m
7/3 m
25/3 m
D = 7/3 + 32/3 + 7/3 = 46/3 = 15.33
m
b- Given : ( a , s ) + v0 , s0
Required: ( s , t )
Method of solution : substitute a = vdv / ds
in the relation ( a , s ), the first
integration step will result in ( v , a ), then substitute v = ds / dt in the relation
( v , s ) and integrate again , the relation ( s , t ) will be obtained.
7
x
c: Given : ( a , v ) + v0 , s0
Required : ( s , t )
Solution : substitute a = dv/dt or a = vdv/ds according to the required relations
and integrate twice to obtain ( s , t ).
EXAMPLE ( 3 ) :
Prob.12.19 (Hibbeler)
The acceleration of a rocket traveling upward is given by
𝒂 = (𝟔 + 𝟎. 𝟎𝟐𝒚) 𝒎/𝒔𝟐 , where 𝑦 is in meters. Determine
the rocket’s velocity when 𝒚 = 𝟐 𝒌𝒎 and the time needed
to reach this attitude. Initially, 𝒗 = 𝟎 and 𝒚 = 𝟎 when
𝒕 = 𝟎.
𝒚
SOLUTION:
𝒗𝒅𝒗
= (𝟔 + 𝟎. 𝟎𝟐𝒚) … … … … … … … … … … … … . . (𝟏)
𝒅𝒚
𝒗
𝒚
𝒗𝟐
∫ 𝒗𝒅𝒗 = ∫ (𝟔 + 𝟎. 𝟎𝟐𝒚) 𝒅𝒚 ∴
= 𝟔𝒚 + 𝟎. 𝟎𝟏 𝒚𝟐
𝟐
𝟎
𝟎
𝒂 =
𝒗 = √𝟏𝟐𝒚 + 𝟎. 𝟎𝟐 𝒚𝟐
= 𝟎. 𝟏√𝟐 √ 𝒚𝟐 + 𝟔𝟎𝟎𝒚 … … … . . (𝟐)
At 𝒚 = 𝟐 𝒌𝒎 = 𝟐𝟎𝟎𝟎 𝒎 in (2): 𝒗 = 𝟎. 𝟏√𝟐 √ (𝟐𝟎𝟎𝟎)𝟐 + 𝟔𝟎𝟎(𝟐𝟎𝟎𝟎) = 𝟑𝟐𝟐. 𝟒𝟗 𝒎/𝒔
𝒗 = 𝟎. 𝟏√𝟐 √ 𝒚𝟐 + 𝟔𝟎𝟎𝒚 = 𝟎. 𝟏√𝟐 √(𝒚𝟐 + 𝟔𝟎𝟎 𝒚 + (𝟑𝟎𝟎)𝟐 ) − (𝟑𝟎𝟎)𝟐
𝒅𝒚
𝒗 = 𝒅𝒕 = 𝟎. 𝟏√𝟐 √(𝒚 + 𝟑𝟎𝟎)𝟐 − (𝟑𝟎𝟎)𝟐
𝒕
𝟐𝟎𝟎𝟎
∫𝟎 𝒅𝒕 = ∫𝟎
𝒅𝒚
(𝟎.𝟏√𝟐 √(𝒚+𝟑𝟎𝟎)𝟐 −(𝟑𝟎𝟎)𝟐 )
∴ 𝒕 = 𝟓√𝟐 (𝒄𝒐𝒔𝒉−𝟏 (𝒚 + 𝟑𝟎𝟎)/𝟑𝟎𝟎)𝟐𝟎𝟎𝟎
𝟎
𝟐𝟑𝟎𝟎
∴ 𝒕 = 𝟓√𝟐 (𝒄𝒐𝒔𝒉−𝟏 ( 𝟑𝟎𝟎 ) − 𝒄𝒐𝒔𝒉−𝟏 (𝟏))
𝟐𝟑𝟎𝟎
∴ 𝒕 = 𝟓√𝟐𝒄𝒐𝒔𝒉−𝟏 ( 𝟑𝟎𝟎 ) = 𝟏𝟗. 𝟐𝟕 𝐬
8
EXAMPLE ( 4 ) :
A particle is moving along a straight line such that it starts from the origin with
a velocity of 𝟒 𝒎 / 𝒔𝒆𝒄. If it begins to decelerate at the rate of 𝒂 = − 𝟐𝒗 m / s2,
where v is in m / s, determine the distance it travels before it stops.
Solution :
𝒂 = 𝒅𝒗/𝒅𝒕 = − 𝟐𝒗
𝒗 𝒅𝒗
,
𝒕
∫𝟒 𝒗 = - 2 ∫𝟎 𝒅𝒕
,
𝒗 = 𝟒 𝒆−𝟐𝒕
𝒔
𝒅𝒗/𝒗 = − 𝟐𝒅𝒕
ln v – ln 4 = -2t = ln ( v / 4)
𝒗 = 𝒅𝒔/𝒅𝒕 = 𝟒 𝒆−𝟐𝒕
,
𝒕
∫𝟎 𝒅𝒔 = ∫𝟎 𝟒 𝒆−𝟐𝒕 𝒅𝒕 .
∴ 𝒔 = −𝟐 ( 𝒆−𝟐𝒕 − 𝟏)
∴ 𝒔 = 𝟐 ( 𝟏 − 𝒆−𝟐𝒕 )
𝟒 𝒆−𝟐𝒕 = 𝟎
when v = 0 :
∴𝒕→∞
∴ 𝒔 = 𝑫 = 𝟐𝒎
EXAMPLE (5 ):
A particle moves along a straight line with an acceleration
a = - 4s m/s2 where
x is in meters. The initial conditions of the motion are : s0 = 0 , v0 = 4 m/s. Find the
relations ( v , t ) , ( s , t ).
Solution :
𝒗𝒅𝒗
𝒗𝟐
𝟒𝟐
𝒔𝟐
,
𝒗 = ±𝟐√𝟒 − 𝒔𝟐
𝒔
∫𝟒 𝒗𝒅𝒗 = ∫𝟎 −𝟒𝒔𝒅𝒔
𝒗𝟐 = 𝟏𝟔 − 𝟒𝒔𝟐 = 𝟒(𝟒 − 𝒔𝟐 )
,
− 𝟐 = −𝟒 𝟐
𝟐
∵ 𝒗𝟎 = +𝟒𝒎/𝒔
𝒅𝒔
𝒔
,
∴ 𝒗 = 𝟐√𝟒 − 𝒔𝟐 = 𝒅𝒕
𝒔
𝒗
,
𝒂 = 𝒅𝒔 = −𝟒𝒔
∫𝟎 √
𝒔𝒊𝒏−𝟏 ( 𝟐) = 𝟐𝒕
,
𝒗 = 𝟒 𝒄𝒐𝒔 𝟐 𝒕
…………. (v , t )
𝒕
𝒅𝒔
𝟒−𝒔𝟐
= 𝟐 ∫𝟎 𝒅𝒕
𝒔 = 𝟐 𝒔𝒊𝒏 𝟐 𝒕
9
………… ( x , t )
EXAMPLE (6):Hibbeler (Example 12-5)
Starting from the position s = +4 m, a particle moving in a straight line has
v = 4 t ( t − 1 ) (t − 2 ) m / s , where
t is the time in seconds. Determine the distance
traveled by the particle:
a) during the first 1.5 seconds. b) during the first 3 seconds.
SOLUTION
𝒗 = 𝟒𝒕(𝒕 − 𝟏)(𝒕 − 𝟐)
∴ 𝒗 = 𝟒𝒕𝟑 − 𝟏𝟐𝒕𝟐 + 𝟖𝒕. . . . . . . . . . . . . . . . . . (𝟏)
With initial conditions :
𝒕 = 𝟎 → 𝒔 = +𝟒𝒎. . . . . . . . . . . . . . . . . . . . . . . (𝟐)
∵𝒗=
𝒔
𝒅𝒔
= 𝟒𝒕𝟑 − 𝟏𝟐𝒕𝟐 + 𝟖𝒕
𝒅𝒕
𝒕
∴ ∫ 𝒅𝒔 = ∫ (𝟒𝒕𝟑 − 𝟏𝟐𝒕𝟐 + 𝟖𝒕)𝒅𝒕 ⇒∴ 𝒔 − 𝟒 = 𝒕𝟒 − 𝟒𝒕𝟑 + 𝟒𝒕𝟐 ⇒
𝟒
𝟎
∴ 𝒔 = 𝒕𝟒 − 𝟒𝒕𝟑 + 𝟒𝒕𝟐 + 𝟒. . . . . . . . . . . . . . (𝟑)
𝒂=
a velocity
𝒅𝒗
= 𝟏𝟐𝒕𝟐 − 𝟐𝟒𝒕 + 𝟖. . . . . . . . . . . . . . . . . . . . . . . (𝟒)
𝒅𝒕
at 𝒗 = 𝟎 → in(𝟏):
∴ 𝒕 = 𝟎 (𝒊𝒏𝒊𝒕𝒊𝒂𝒍 𝒎𝒐𝒎𝒆𝒏𝒕), 𝒕 = 𝟏 𝒔𝒆𝒄 , 𝒕 = 𝟐 𝒔𝒆𝒄
∴ 𝒔𝟏 = (𝟏)𝟒 − 𝟒(𝟏)𝟑 + 𝟒(𝟏)𝟐 + 𝟒 = 𝟓𝒎,
𝒔𝟐 = (𝟐)𝟒 − 𝟒(𝟐)𝟑 + 𝟒(𝟐)𝟐 + 𝟒 = 𝟒𝒎
𝒂𝟏 = 𝟏𝟐(𝟏)𝟐 − 𝟐𝟒(𝟏) + 𝟖 = −𝟒𝒎/𝒔𝟐 (reversetoleft ←),
𝒂𝟐 = 𝟏𝟐(𝟐)𝟐 − 𝟐𝟒(𝟐) + 𝟖 = +𝟖𝒎/𝒔𝟐 (reversetoright →
10
at: 𝒕 = 𝟏. 𝟓 𝒔𝒆𝒄 ⇒ 𝒊𝒏(𝟑):
𝒔𝟏.𝟓 = (𝟏. 𝟓)𝟒 − 𝟒(𝟏. 𝟓)𝟑 + 𝟒(𝟏. 𝟓)𝟐 + 𝟒
𝒔𝟏.𝟓 = 𝟒. 𝟓𝟔𝒎,
at: 𝒕 = 𝟑 𝒔𝒆𝒄 ⇒ 𝒊𝒏(𝟑):
𝒔𝟑 = (𝟑)𝟒 − 𝟒(𝟑)𝟑 + 𝟒(𝟑)𝟐 + 𝟒
𝒔𝟑 = 𝟏𝟑𝒎
𝑫(𝑻=𝟎→𝑻=𝟏.𝟓) = (𝟓 − 𝟒) + (𝟓 − 𝟒. 𝟓𝟔) = 𝟏. 𝟒𝟒𝒎,
𝑫(𝑻=𝟎→𝑻=𝟑 = (𝟓 − 𝟒) + (𝟓 − 𝟒) + (𝟏𝟑 − 𝟒) = 𝟏𝟏𝒎
)
From the path plot showing the positions of the particle at the different instants of
time, the distance traveled by the particle during the first
1.5 seconds and during
the first 3 seconds can be calculated as follows:
O
t=0 ,
t=1s
A
B
t=3 s
𝒔
C
t=2s
D
E
4m
5m
13 m
4.56 m
t=1.5s
11
EXAMPLE (7):
A particle moves along a straight line according to the relation between the
position s and the time t given by : 𝒔 = 𝟐 𝒕𝟑 − 𝟐𝟒 𝒕 + 𝟔 meters .Determine :
1. the time required till a velocity 𝒗 = 𝟕𝟐 𝒎/𝒔 can be attained
2. the acceleration when 𝒗 = 𝟑𝟎𝒎/𝒔
3. the distance traveled in the time interval t = 1 → t = 4 sec
SOLUTION
𝑠 = 2𝑡 3 − 24 𝑡 + 6. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . … … … … . (1)
𝑑𝑠
= 6𝑡 2 − 24. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
𝑑𝑡
𝑑𝑣
𝑎=
= 12𝑡. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . … … (3)
𝑑𝑡
𝑣=
1. At 𝑣 = 72 𝑚/𝑠 in (2) :
72 = 6𝑡 2 − 24 ∴ 𝑡 2 = 16,
𝑡 = 4𝑠𝑒𝑐
2. At 𝑣 = 30𝑚/𝑠 in (2) : ∴ 30 = 6𝑡 2 − 24 ∴ 𝑡 2 = 9, 𝑡 = 3 𝑠𝑒𝑐
in (3):
𝑎 = 12(3) = 36𝑚/𝑠 2
3. The distance traveled in the time interval 𝑡 = 1 → 𝑡 = 4𝑠𝑒𝑐 :
at v = 0 : in (2):
0 = 6𝑡 2 − 24 ∴ 𝑡 = 2 𝑠𝑒𝑐,
𝑎 = 12(2) = +24𝑚/𝑠 2 (𝑟𝑒𝑣𝑒𝑟𝑠𝑒𝑡𝑜𝑡ℎ𝑒 + 𝑣𝑒𝑑𝑖𝑟𝑒𝑐𝑡𝑖𝑜𝑛)
𝑎𝑡 𝑡 = 1:
𝑎𝑡 𝑡 = 2:
at 𝑡 = 4:
𝑠 = 2(1)3 - 24(1) + 6 = −16 𝑚
𝑠 = 2(2)3 - 24(2) + 6 = −26 𝑚
𝑠 = 2(4)3 - 24(4) + 6 = +𝟑𝟖 𝑚
𝐷( t=1→𝑡=4) = 𝐴𝐵 + 𝐵𝐶 = (26 − 16) + (26 + 𝟑𝟖) = 𝟕𝟒𝒎
12
t = 2s
t = 1s
t = 3s
(26-16)=10
16m
26m
B
A
O
C
38 m
26 + 38 = 64 m
13
Constant Acceleration, a = ac.
When the acceleration is constant, each of the three kinematic equations 𝑎𝑐 = 𝑑𝑣/𝑑𝑡,
𝑣 = 𝑑𝑠/𝑑𝑡, and 𝑎𝑐 = 𝑣 𝑑𝑣/𝑑𝑠 can be integrated to obtain formulas that relate
𝑎𝑐 , 𝑣, 𝑠, and 𝑡.
Velocity as a Function of Time.
Integrate 𝑎𝑐 = 𝑑𝑣/𝑑𝑡, assuming that initially 𝑣 = 𝑣𝑜 , when 𝑡 = 0.
𝑣
𝑡
∫ 𝑑𝑣 = ∫ 𝑎𝑐 𝑑𝑡
𝑣0
0
𝒗 = 𝒗𝟎 + 𝒂𝒄 𝒕 … … … … … . (1)
𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒂𝒄𝒄𝒆𝒍𝒆𝒓𝒂𝒕𝒊𝒐𝒏
Position as a Function of Time.
Integrate 𝑣 = 𝑑𝑠/𝑑𝑡 = 𝑣0 + 𝑎𝑐 𝑡, assuming that initially 𝑠 = 𝑠𝑜 when 𝑡 = 0.
𝑠
𝑡
∫ 𝑑𝑠 = ∫ (𝑣0 + 𝑎𝑐 𝑡) 𝑑𝑡
𝑠0
0
𝒔 = 𝒔𝟎 + 𝒗𝟎 𝒕 +
𝟏
𝒂 𝒕𝟐 … … … (𝟐)
𝟐 𝒄
𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒂𝒄𝒄𝒆𝒍𝒆𝒓𝒂𝒕𝒊𝒐𝒏
Velocity as a Function of Position.
Either solve for 𝑡 in (1) and substitute into (2), or integrate 𝑣 𝑑𝑣 = 𝑎𝑐 𝑑𝑠, assuming
that initially 𝑣 = 𝑣0 at 𝑠 = 𝑠0.
𝑣
𝑠
∫ 𝑣 𝑑𝑣 = ∫ 𝑎𝑐 𝑑𝑠
𝑣0
𝑠0
𝒗𝟐 = 𝒗𝟐𝒐 + 𝟐𝒂𝒄 (𝒔 − 𝒔𝟎 ) … … … (3)
𝑪𝒐𝒏𝒔𝒕𝒂𝒏𝒕 𝒂𝒄𝒄𝒆𝒍𝒆𝒓𝒂𝒕𝒊𝒐𝒏
Example 12.3 P Hibbeler (page:12)
During a test a rocket travels upward at 75 𝑚/𝑠, and when it is 40 𝑚 from the ground its engine fails.
Determine the maximum height 𝑠𝐵 reached by the rocket and its speed just before it hits the ground.
While in motion the rocket is subjected to a constant downward acceleration of 9.81 𝑚/𝑠 2 due to
gravity. Neglect the effect of air resistance.
14
15
Example : Hibbeler (12-31):
16
Rectilinear motion-Hibbeler additional problems
Prob.12.8:
A particle travels along a straight line with a velocity 𝒗 = (𝟏𝟐 − 𝟑𝒕𝟐 ) 𝒎/𝒔,
where t is in seconds. 𝑾𝒉𝒆𝒏 𝒕 = 𝟏 𝒔, the particle is located 𝟏𝟎 𝒎 to the left of the
origin. Determine the acceleration when 𝒕 = 𝟒 𝒔, the displacement from 𝒕 = 𝟎
to
𝒕 = 𝟏𝟎 s, and the distance the particle travels during this time period.
SOLUTION:
𝒗 = (𝟏𝟐 − 𝟑𝒕𝟐 ) 𝒎/𝒔 … … … … … … . . (𝟏) ,
𝒂=
𝒅𝒗
= −𝟔𝒕
𝒅𝒕
𝒎
… … … … … … … … . (𝟐) ,
𝒔𝟐
𝒗 =
𝒕
𝒕
𝒅𝒙
= (𝟏𝟐 − 𝟑𝒕𝟐 ) ∴ ∫ 𝒅𝒙 = ∫ (𝟏𝟐 − 𝟑𝒕𝟐 ) 𝒅𝒕
𝒅𝒕
−𝟏𝟎
𝟏
∴ 𝒙 + 𝟏𝟎 = (𝟏𝟐𝒕 − 𝒕𝟑 ) − (𝟏𝟏)
∴ 𝒙 = −𝟐𝟏 + 𝟏𝟐 𝒕 − 𝒕𝟑 … … . . … … … . (𝟑)
From (2) : 𝒂|𝒕=𝟒 = −𝟔(𝟒) = −𝟐𝟒 𝒎/𝒔𝟐
The displacement from 𝒕 = 𝟎 to
travels during this time period:
𝒕 = 𝟏𝟎 s, and the distance the particle
From (1) if 𝒗 = 𝟎 = (𝟏𝟐 − 𝟑𝒕𝟐 ), then 𝒕 = 𝟐 𝒔𝒆𝒄 , then in (3) :
𝒙|𝒕=𝟎 = −𝟐𝟏 𝒎
,
𝒙|𝒕=𝟐 = −𝟐𝟏 + 𝟏𝟐(𝟐) − (𝟐)𝟑 = −𝟓 𝒎
,
𝒙|𝒕=𝟏𝟎 = −𝟐𝟏 + 𝟏𝟐(𝟏𝟎) − (𝟏𝟎)𝟑 = −𝟗𝟎𝟏 𝒎
∆𝒙 = −(𝟗𝟎𝟏 − 𝟐𝟏) = −𝟖𝟖𝟎 𝒎
𝑫 = (𝟐𝟏 − 𝟓) + (𝟗𝟎𝟏 − 𝟓) = 𝟗𝟏𝟐 𝒎
𝒕=𝟎
𝟐𝟏 𝒎
∆𝒙
𝑪
𝑨
𝑶
𝑩
𝟓𝒎
s𝒕 = 𝟐
𝟗𝟎𝟏 𝒎
𝒕 = 𝟏𝟎 𝒔
17
𝒙
Prob.12.9:
When two cars A and B are next to one another, they are traveling in the same
direction with speeds 𝒗𝑨 and 𝒗𝑩 respectively. If B maintains its constant speed,
while A begins to decelerate at 𝒂𝑨 determine the distance d between the cars at
the instant A stops.
,
SOLUTION:
18