Faculty of EMS
Department: Mechatronics
Mechanics II
Prof. Dr. E. I. Morgan
Tutorial (1)
Subject: Rectilinear Motion
Prob.1: [11.1]
The motion of a particle is defined by the relation:
x = 8t3 – 8 + 30 sin(πt),
where x and t are expressed in millimeters and seconds, respectively. Determine the
position, velocity and acceleration of the particle when t = 5 sec.
Answer:
x(5)= 992 mm
v(5)= 505.75 mm/s
a(5)= 240 mm/s2
-1-
Prob.2: [11.3]
The motion of the slider A is defined
by the relation x = 20 sin(kt),
where x and t are expressed in cm and
seconds, respectively, and k is a
constant. Knowing that k =10 rad/sec,
determine the position, velocity and
acceleration of the slider A when
t = 0.05 s.
Answer:
x(0.05) = 9.59 cm
v(0.05) = 175.5 cm/s
a(0.05) = -959 cm/s2
-2-
Prob.3: [11.8]
The motion of a particle is defined by the relation
x =2t3 ‒ 18t2 + 48t ‒16,
where x and t are expressed in millimeters and seconds, respectively.
Determine:
a) when the velocity is zero,
b) the position and the total distance traveled when the acceleration is zero,
c) the displacement, total distance traveled, average velocity during the time
interval 0 ≤ t ≤ 6 seconds. Sketch the motion during the mentioned interval.
Answer:
a) Time = 4 sec., 2 sec.
b) Position: 20 mm, Distance: 44mm
-3-
Prob.4: [11.9]
The acceleration of point A is defined by the
relation a = ‒ 1.8 sin (kt), where a and t are
expressed in m/s2 and seconds, respectively,
and k = 3 rad/s. knowing that x = 0 and
v = 0.6 m/s when t = 0, determine the velocity
and position of point A when t = 0.5 sec.
Answer:
v(0.5) = 0.0424 m/s
x(0.5) = 0.199 m
-4-
Prob.5: [11.12]
It is known that from t =2s to t =10s the acceleration of a particle is inversely
proportional to the cube of the time t. When t = 2s, v = - 15 cm/s, and when
t = 10s, v = 0.36 cm/s. Knowing that the particle is twice as far from the
origin when t = 2s as it is when t = 10s, determine;
a) the position of the particle when t = 2s and when t = 10s,
b) the total distance traveled by the particle from t =2s to t = 10s.
Answer:
a) x(2) = 35.2 cm, x(10) = 17.6 cm
b) distance (xT) = 18.4 cm
-5-
Prob.6: [11.18]
Point A oscillates with an acceleration a = 144 (20 ‒ x), where a and x are
expressed in cm/s2 and cm respectively. Knowing that the system starts at time
t = 0 with v = 0 and x = 19 cm, determine the position and velocity of A when
t = 0.2 s.
Answer:
x(0.2) = 20.7 cm
v(0.2) = 8.11 cm/s
-6-
Prob.7: [11.21]
The acceleration of a particle is defined by the relation a = - k v , where k is a
constant. Knowing that x = 0 and v = 25 cm/s at t= 0, and that v = 12 cm/s when
x = 6 cm, determine;
a) the velocity of the particle when x =8cm,
b) the time required for the particle to come to rest.
Answer:
a) v = 5.74 cm/s
b) t = 1.079 s
-7-
Prob.8: [11.33]
An airplane begins its take-off run at A with zero velocity and a constant
acceleration a. knowing that it becomes airborne 30 s later at B and that the
distance AB is 900 m, determine:
(a) the acceleration a,
(b) the take-off velocity vB.
Answer:
a) a = 2 m/s2
b) vB = 60 m/s
-8-
Prob.9: [11.49]
Block C starts from rest and moves downward with a constant acceleration.
Knowing that after 12 s the velocity of block A is 456 mm/s, determine;
a) the acceleration of A, B and C,
b) the velocity and the change in position of block B after 8 s.
Answer:
a) aA = 38 mm/s2, aB = -25.33 mm/s2, aC = 12.67 mm/s2
b) v = -202.64 mm/s , ∆s = -810.56 mm
-9-
Prob. 10: (M.T. Spring 2017)
Three blocks A, C, D, and collar B are connected
using two cables and massless pulleys as shown.
The system is released from rest at t = 0,and Block
A moves upward with a constant acceleration of
aA= 120 mm/s2 while block C moves downward
with constant acceleration of aC = 40 mm/s2
(a) Deduce the kinematic relation(s)
between the four blocks. State the
number of Degrees of freedom of the
system.
(b) Find the corresponding accelerations of
collar B, and block D.
(c) Determine the velocity of B after 3
seconds, and the distance covered by
block D during the first 5 seconds.
(d) Find the relative velocity of block B
with respect to D after 3 seconds.
- 10 -
xA
xD
xB
xC