1. Graph of velocity (v) vs. time (t) shown in the figure below. What is the deceleration according to the graph.
Solution
A-B = motion at constant acceleration, B-C = motion at constant velocity, C-D = motion at constant deceleration.
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2. Graph of linear motion shown in the figure below. What is the distance traveled by an object from 0 second – 8 seconds.
Solution
Area 1 = triangle area = ½ (4-0)(12-0) = ½ (4)(12) = (2)(12) = 24
Area 2 = rectangle area = (8-4)(12-0) = (4)(12) = 48
Distance traveled during 8 seconds = 24 meters + 48 meters = 72 meters.
3. Graph of velocity (v) vs. time (t) for linear motion shown in figure below. What is the distance traveled during 12 seconds.
Solution
Area 1 = area of triangle = ½ (2-0)(4-0) = ½ (2)(4) = 4
Area 2 = area of rectangle = (6-2)(4-0) = (4)(4) = 16
Area 3 = area of triangle = ½ (8-6)(4-0) = ½ (2)(4) = 4
Area 4 = area of triangle = ½ (10-8)(4-0) = ½ (2)(4) = 4
Area 5 = area of square = (12-10)(4-0) = (2)(4) = 8
The distance traveled during 12 seconds = 4 + 16 + 4 + 4 + 8 = 36 meters
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4. An object travels at a constant 36 km/hour during 5 seconds, then accelerated 1 m/s2 during 10 seconds and then decelerated 2 m/s2 until rest. Which graph (v-t) shows the object’s travels.
Known :
Motion 1 = constant velocity
Constant velocity (v) = 36 km/jam = 36 (1000 m) / 3600 s = 36,000 m / 3600 s = 10 m/s
Time interval (t) = 5 seconds
Motion 2 = constant acceleration
Initial velocity (vo) = velocity in motion 1 = 10 m/s
Acceleration (a) = 1 m/s2
Time interval (t) = 10 seconds
Motion 3 = constant deceleration
Deceleration (a) = -2 m/s2
Final velocity (vt) = 0 m/s
Wanted : which graph shown object’s travels
Solution :
The final velocity of motion 2 :
vt = vo + a t
vt = 10 + (1)(10) = 10 + 10 = 20 m/s
Final velocity of motion 2 (vt) = initial velocity of motion 3 (vo) = 20 m/s
Time interval of motion 3 :
vt = vo + a t
0 = 20 + (-2)(t)
0 = 20 – 2t
20 = 2t
t = 20/2
t = 10 seconds
Motion 1 : Object travels at a constant 10 m/s in 5 seconds
Motion 2 : Object accelerated in 10 seconds until it’s speed = 20 m/s.
Motion 3: Object decelerated in 10 seconds until rest.
Graph B shows the object’s travels.
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5. A car travels at a constant 5 m/s in 10 seconds, then accelerated 1 m/s2 in 5 seconds. Then decelerated until rest after car travels 137.5 meters. Which graph (v-t) shows the object’s travels.
Known :
Motion 1 = motion at constant velocity
Constant velocity (v) = 5 m/s
Time interval (t) = 10 seconds
Motion 2 = motion at constant acceleration
Initial velocity (vo) = velocity in motion 1 = 5 m/s
Acceleration (a) = 1 m/s2
Time interval (t) = 5 seconds
Final velocity (vt) = 0 m/s
Total distance (s) = 137.5 m/s
Wanted : Graph shows car’s travel
Solution :
Distance for motion 1 :
s = v t = (5)(10) = 50 meters
Distance for motion 2 :
s = vo t + ½ a t2 = (5)(5) + ½ (1)(5)2 = 25 + ½ (25) = 25 + 12.5 = 37.5 meters
Distance for motion 3 :
137.5 – (50 + 37.5) = 137.5 – 87.5 = 50 meters
The final velocity of motion 2 :
The final velocity of motion 2 calculated using data of motion 2 :
vt = vo + a t
vt = 5 + (1)(5) = 5 + 5 = 10 m/s
Final velocity for motion 2 (vt) = initial velocity for motion 3 (vo) = 10 m/s
Time interval for motion 3 :
Time interval calculated after find deceleration. Deceleration (a) and time interval (t) calculated using data of motion 3.
Deceleration of object in motion 3 :
vt2 = vo2 + 2 a d
02 = 102 + 2 a (50)
0 = 100 + 100 a
100 = -100 a
a = – 100 / 100
a = – 1 m/s2
Time interval of motion 3 :
vt = vo + a t
0 = 10 + (-1) t
0 = 10 – t
t = 10 seconds
Motion 1 : Object travels at a constant 5 m/s in 10 seconds.
Motion 2 : Object travels during 5 seconds until it’s velocity = 10 m/s.
Motion 3: Object decelerated for 10 seconds until rest.
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6. An 800-kg car travels along a straight line with the initial velocity of 36 km/hour. After travels 150 meters, car’s velocity = 72 km/hour. Determine the time interval.
Known :
vo = 36 km/hour = 36,000 meters / 3600 s = 10 m/s
vt = 72 km/hour = 72,000 meters / 3600 s = 20 m/s
d = 150 meters
Wanted : time interval (t)
Solution :
Three equation of motion at constant acceleration :
vt = vo + a t
d = vo t + ½ a t2
vt2 = vo2 + 2 a d
Acceleration :
vt2 = vo2 + 2 a d
202 = 102 + 2 a (150)
400 = 100 + 300 a
400 – 100 = 300 a
300 = 300 a
a = 300 / 300
a = 1 m/s2
Time interval :
vt = vo + a t
20 = 10 + (1) t
20 – 10 = t
t = 10 seconds
[irp]
7. A car travels at a constant 20 m/s. Car rest in 5 seconds after decelerated. What is the distance traveled by car.
Known :
Initial speed (vo) = 20 m/s
Final speed (vt) = 0 m/s
Time interval (t) = 5 seconds
Wanted : distance (d)
Solution :
deceleration :
vt = vo + a t
0 = 20 + a 5
-20 = 5a
a = -20/5
a = -4 m/s2
Distance traveled by car :
d = vo t + 1/2 a t2
d = (20)(5) + 1/2 (-4)(5)2 = (20)(5) + (-2)(25)
d = 100 – 50
d = 50 meters
8. Distance and time interval of a moving object shown in table below.
Determine the types of motion experienced by objects 1 and 2…
Solution
Object 1 :
v = s / t
v = velocity, s = distance, t = time interval
v = 60/15 = 4 cm/s
v = 80/20 = 4 cm/s
v = 100/25 = 4 cm/s
Constant velocity = uniform linear motion.
Object 2 :
v = s / t
v = velocity, s = distance, t = time interval
v = 4/2 = 2 cm/s
v = 9/3 = 3 cm/s
v = 25/5 = 5 cm/s
Velocity increases = nonuniform linear motion, accelerated.
