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Typology: Exercises
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Newton’s Law of Motion for a Particle
Free falling Bodies, Air resistance is Neglected.
Sample Problems
A C
D
S
330
𝑚 𝑠 (^) S
Solution:
T1 = time for the ball to travel a distance S
T2 = time for the sound to travel a distance S
T1 = √2s/g
T2 = S/
Substitute in equation 1:
√2S/9.81 + S/330 = 5
0.452√S + S/330 – 5 = 0
S = 149√S – 1650 = 0
Let y = √S
y² = S
y² + 149y -1650 = 0
y = 10.
√S = 10.
S = 107.2 m
t = 1.4 sec.
S = 15(1.4) – 4.905(1.4)²
S = 11.4 m. from theground
3.A ball is thrown vertically upward with an initial velocity of 3 m/s from the window of a tall building. The ball strikes the sidewalk at the ground level 4 seconds later. Determine the velocity with which the ball hits the ground and the height of the window above the ground level.
Solution:
Vf² = Vo² - 2gh
= (3)² - 2(9.81)h
h = 0.459 m.
Vf = Vo – gt
0 = 3 – 9.81(t)
h
t = 0.
t2 = 4 secs. – 0.31 secs.
t2 = 3.69 secs.
H = ½ g(t2)²
H = ½ (9.81)(3.61)
H = 66.79 m
Height of the window = H – h
S = 66.79 – 0.459 = 66.
V3² = V1² + 2gh
V3 = 0 + 2(9.81)(66.79)
V3 = 36.20 m/s
Further Application of Kinematics of Translation
Sample Problems
3.the acceleration of a particle is given by 3 when t=0,s=-40m, and
v= 30. Determine the position and velocity after t=2 sec.
3
Projectile Motion
Solution:
.
.
.3 30 (eq.2)
Eq1 to eq. 2
Given x=60ft.
Θ=60° y=3ft.
Solution
5 ft.
60 ft.
10 ft. 60 °
1.) A train is to commute between station A and station B with a top speed of 250 kph but cannot accelerate nor decelerate faster than 4 m/sec². what is the minimum distance between the two stations in order for the train to be able to reach its top speed? 0 000 3 00 .
Accelearation = 15(60) = 900km/hr²
Deceleration = 10(60) = 600 km/hr²
900t1 = 60
t1 = = 4 min
600t3 = 60
t3 = = 6 min.
S1= (t1) =
S1 = 2 km
S2 – S1 = 60t
S2 – 2 = 60t
10 – S2 = (60/2)(t3)
S2 = 10 – 3 = 7 km
S2 – S1 = 7 – 2 = 5 km
5 = 60t
t2 = 5(60)/60 = 5 min.
total time: 6 + 4 + 5 = 15 mins
VELOCITY DIAGRAM
1.8t1 = V eq. 1 1.8t3 = V eq. 2
From eq. (1) and (2) 1.8t1 = 1.8t
t1 = t3 eq. 3
t1 + t2 + t3 = 40
2t1 + t = 40
t2 = 40 – 2t1 eq. 4
1.8 m/s ACCELERATION DIAGRAM
VELOCITY DIAGRAM
t 1
t 1
t 2
t 2
t 3
t 3
S 1
S 2
S 3 = 540
DISTANCE DIAGRAM
2 °
1 ° 2 °
V V
Rectilinear Motion with Constant Acceleration.
Given: a=-2ft/
3 =-2(2)S
S= 1936ft
Given: t=10sec
Given: 0 1 1 1 1 (^1) 3. 1 (1)
