1.
The table above shows data of three objects that travel the same distance at constant acceleration.
What are the final speed of object P and the initial speed of object Q?
Solution :
First, determine the distance traveled by object 3.
Distance traveled by object 3 :
Known :
Initial speed (vo) = 0 m/s
Final speed (vt) = 30 m/s
Acceleration (a) = 3 m/s2
Wanted : Distance
Solution :
vt2 = vo2 + 2 a s
vt2 – vo2 = 2 a s
302 – 02 = 2 (3) s
900 – 0 = 6 s
900 = 6 s
s = 900 / 6
s = 150 meters
The final velocity of object 1 :
Known :
Initial speed (vo) = 20 m/s
Acceleration (a) = 4 m/s2
Distance (s) = 150 meters
Wanted : Final speed (vt)
Solution :
vt2 = vo2 + 2 a s
vt2 = 202 + 2 (4)(150)
vt2 = 400 + 1200
vt2 = 1600
vt = 40 m/s
The initial velocity of object 2 :
Known :
Final speed (vt) = 50 m/s
Acceleration (a) = 3 m/s2
Distance (s) = 150 meters
Wanted : Initial speed (vo)
Solution :
vt2 = vo2 + 2 a s
vt2 – 2 a s = vo2
502 – 2(3)(150) = vo2
2500 – 900 = vo2
1600 = vo2
vo = 40 m/s
[irp]
2. Three objects travel on a horizontal plane at constant acceleration. The three objects have the same acceleration. Data of the three object when travels in 10 seconds, shown in figure below.
Determine P and Q.
Solution :
First, determine the acceleration of object 1.
Acceleration of object 1 :
Known :
The initial speed (vo) = 2 m/s
The final speed (vt) = 22 m/s
Distance (s) = 120 meters
Wanted : Distance
Solution :
vt2 = vo2 + 2 a s
vt2 – vo2 = 2 a s
222 – 22 = 2 a (120)
484 – 4 = 240 a
480 = 240 a
a = 480/240
a = 2 m/s2
The initial speed of object 2 :
Known :
Acceleration (a) = 2 m/s2
Final speed (vt) = 24 m/s
Distance (s) = 140 meters
Wanted : Initial speed (vo)
Solution :
vt2 = vo2 + 2 a s
242 = vo2 + 2 (2)(140)
576 = vo2 + 560
576 – 560 = vo2
16 = vo2
vo = 4 m/s
Distance of object 3 :
Known :
Initial speed (vo) = 0 m/s
Final speed (vt) = 20 m/s
Acceleration (a) = 2 m/s2
Wanted : Distance (s)
Solution :
vt2 = vo2 + 2 a s
202 = 02 + 2 (2) s
202 = 2 (2) s
400 = 4 s
s = 400/4
s = 100 meters
[irp]
3. Determine the distance traveled by object in 40 seconds.
Solution :
Area 1 = area of rectangle = (20-0)(8-0) = (20)(8) = 160 meters
Area 2 = area of triangle = ½ (25-20)(8-0) = ½ (5)(8) = (5)(4) = 20 meters
Area 3 = area of triangle = ½ (30-25)(8-0) = ½ (5)(8) = (5)(4) = 20 meters
Area 4 = area of rectangle = (40-30)(8-0) = (10)(8) = 80 meters
The distance traveled in 40 seconds = 160 + 20 + 20 + 80 = 280 meters
4. The change of object’s speed in 2 seconds stated by graph below. Determine distance traveled by the object.
Solution :
Area 1 = area of triangle = ½ (5-0)(20-0) = ½ (5)(20) = (5)(10) = 50 meters
Area 2 = area of rectangle = (15-5)(20-0) = (10)(20) = 200 meters
Area 3 = area of triangle = ½ (20-15)(20-0) = ½ (5)(20) = (5)(10) = 50 meters
Distance traveled during 20 seconds = 50 + 200 + 50 = 300 meters