Chapter I:
RECTILINEAR
MOTION
PROBLEMS
1. The velocity of a particle which moves along the
s-axis is given by s  40  3t 2 m/s, where t is in seconds.
Calculate the displacement Ds of the particle during
the interval from t = 2 s to t = 4 s. (2/6)
SOLUTION
s  40  3t 2 displacement Ds from t = 2 s to t = 4 s.
s  v  40  3t
,
ds
v
dt
,
ds  vdt
 40  3t dt
s
t 4
0
t 2
 ds 
2
2
3 t
s  40 t 
3
3
4
 40 (4)  43  40 (2)  23
2
160  64  80  8  24 m
Ds  24 m
2. In the pinewood-derby event shown, the car is
released from rest at the starting position A and
then rolls down the incline and onto the finish line
C. If the constant acceleration down the incline is
2.75 m/s2 and the speed from B to C is essentially
constant, determine the time duration tAC for the
race. The effects of the small transition area at B
can be neglected. (2/14)
SOLUTION
A B
time duration tAC for the race
a AB  cst  2.75 m / s 2 ,
vA  0
1
s AB  s A  v At AB  a ABt AB 2
2
1
1
s AB  a ABt AB 2  3  (2.75 )t AB 2 , t AB  1.477 s
2
2
vdv  ads
vB
,
v B2
3
vdv

2
.
75
ds
,

2
.
75
s


0
2
0
0
3
v B  4.06 m / s ,
B C
a BC 0
,
v B  vC
1
s BC  s B  v B t BC  a BC t BC 2
2
4  4.06 t BC , t BC  0.985 s
ttotal t AB  t BC  1.477  0.985  2.462 s
,
v B2  2(2.75 )(3)  16 .5
3. A particle starts from rest at x = -2 m and moves
along the x -axis with the velocity history shown. Plot
the corresponding acceleration and the displacement
histories for the 2 seconds. Find the time t when the
particle crosses the origin. (2/29)
SOLUTION
time t when the
particle crosses
the origin = 0.917 s.
0  0.5 s
t, s
a, m/s2
6
v
t, s
0
-4
0  1.25  3t
t s 0  0.417 s
ttotal0.5 0.4170.917 s
v
x, m
0.97
s  s 0  v0 t
1.375
from 0.5 s
1.250
1 2
s  s0  v 0 t  at
2
1
s  2  (6)( 0.5) 2
2
st 0.5  1.25 m
0.5
t, s
0
1
-1.25
-2
1.5
1.75 2
4. A sprinter reaches his maximum speed vmax in 2.5
seconds from rest with constant acceleration. He
then maintains that speed and finishes the 100
meters in the overall time of 10.40 seconds.
Determine his maximum speed vmax. (2/32)
t = 10.40 s
SOLUTION
v0=0, a=cst, vmax in 2.5 s, then a=0, stotal=100 m, ttotal=10.40 s, vmax=?
first 2.5 s
vmax  v 0  at  2.5a
s  s 0  v 0t 
1 at 2  1 a(2.5) 2  3.125 a
2
2
last 7.9 s
s  s0  v0t  12 a t 2  3.125 a  2.5a (7.9)
 3.125 a  19 .75 a  22 .875 a
100  22 .875 a
a  4.37 m / s 2
vmax  10 .93 m / s
t = 10.40 s
5. The cone falling with a speed v0 strikes and penetrates
the block of packing materials. The acceleration of the cone
after impact is a = g – cy2, where c is a positive constant and
y is the penetration distance. If the maximum penetration
depth is observed to be ym, determine the constant c. (2/45)
SOLUTION
vinitial  v0 , a  g  cy 2  f ( y ) ,
v final  0 , ymax  ym , c  ?
vdv  ady
 g  cy dy
0
ym
v0
0
 vdv 
2
3
cym
1 2
 v0  gym 
2
3
,
3 2 6
 v0  gym
2
2
,
3
cym
3
2
cy
m
v02 
 2 gym
3
3  v02  2 gym 
c
3

2 
ym

6. A test projectile is fired horizontally into a viscous
liquid with a velocity v0. The retarding force is
proportional to the square of the velocity, so that the
acceleration becomes a = -kv2. Derive expressions for the
distance D traveled in the liquid and the corresponding
time t required to reduce the velocity to v0 /2. Neglect
any vertical motion. (2/54)
SOLUTION
vinitial =v0 , a = -kv2 , derive expressions for the distance D traveled in the liquid and the
corresponding time t required to reduce the velocity to v0 /2.
vdv  adx
,
vdv
 dx ,
2
 kv
a  f (v )
v0 / 2

v0
D
dv
  k  dx
v
0
 v0 / 2 
1

  ln    0.693   kD
ln 
2
 v0 
0.693
D
k
SOLUTION
vinitial =v0 , a = -kv2 , derive expressions for the distance D traveled in the liquid and the
corresponding time t required to reduce the velocity to v0 /2.
dv
a
dt
dv
 kv
2
1
v
1
,
a  f (v )
 dt ,
v0 / 2
t
v0
0
2
v
dv  k  dt

v0 / 2
 kt
v0
1
t
kv0
,
dv
 dt
a
,
2
1
1
 

 kt
v0 v0
v0
7. A bumper, consisting of a nest of three springs, is used to
arrest the horizontal motion of a large mass which is traveling
at 40 m/s as it contacts the bumper. The two outer springs
cause a deceleration proportional to the spring deformation.
The center spring increases
the deceleration rate when
the compression exceeds 0.5
m as shown on the graph.
Determine
the
maximum
compression x of the outer
springs. (2/55)
SOLUTION
 vdv   adx
v0=40 m/s , maximum compression x of the outer springs
,
0
x
40
0
 vdv   adx  area under a  x curve
z
2000
2000
x  0.5

, z
x  0.5
0.5
0.5
z
x
2000
SOLUTION
v0=40 m/s , maximum compression x of the outer springs
 vdv   adx
,
0
x
40
0
under a  x curve
 vdv   adx  area
z
1
1 2000
1

2

 40     1000(0.5)  
( x  0.5)( x  0.5)  1000( x  0.5)
2
2 0.5
2


 1600  2 250  2000( x  0.5) 2  1000( x  0.5)
1600  500  4000x 2  x  0.25  2000 x  1000


4000x 2  4000x 1000
4000 x 2  2000 x  1100  0
x 2  0.5 x  0.275  0
x1  0.831 m
x1  0.331 m

8. The preliminary design for a rapid-transit system calls for the
train velocity to vary with time as shown in the plot as the train
runs the 3.2 km between stations A and B. The slopes of the
cubic transition curves (which are of form a+bt+ct2+dt3) are zero
at the end points. Determine the total run time t between the
stations and the maximum acceleration. (2/58)
SOLUTION
slopes of a+bt+ct2+dt3 form curves zero at end points, determine total run time t
between the stations and the maximum acceleration
v  f (t )  a  bt  ct 2  dt 3
initial and final conditions give the constants
v0  0 at t  0
 a0
dv
 a  b  2ct  3dt 2  0
dt
then,
(at t  0 , t  15 s )
2ct  3dt 2  0
2c(15)  3d (15) 2  0
130
30c  675d  0 ,
675d
c
 c  22.5d
30

b0
SOLUTION
vt 15  130 km / h  36 .11 m / s  ct 2  dt 3
 c(15 ) 2  d (15 ) 3  225 c  3375 d
36 .11  225 (22 .5d )  3375 d  1687 .5d

 

5062.5d
130
d  0.0214
m / s 4 , c  0.481 m / s 3
v  f (t )  0.481t 2  0.0214 t 3
dv
a  f (t ) 
 2(0.481)t  3(0.0214 )t 2  0.963 t  0.0642 t 2
dt
the distance the train travels in the first and last 15 seconds


s
15
ds
 v  f (t )  ds   0.481t 2  0.0214 t 3 dt ,
dt
0
0
3
0.481t
0.0214 t
s

3
4
4
15
 271 m
0
SOLUTION
when the speed is constant, the train travels
3200 – 2(271)=2658 m
1 2
s  s 0  v0t  a t , 2658  36 .11t , t  73 .61 s
2
total run time
ttotal  2(15)  73 .61  103 .61 s
130
Maximum acceleration
a  f (t )  0.963 t  0.0642 t
2
da
 0.963  2(0.0642 )t  0 ,
dt
da
0
dt
t  7.5 s
amax,t 7.5  0.963 (7.5)  0.0642 (7.5) 2  3.61 m / s 2