Nonuniform linear motion – problems and solutions

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 (v_o) = 0 m/s

Final speed (v_t) = 30 m/s

Acceleration (a) = 3 m/s²

Wanted : Distance

Solution :

v_t² = v_o² + 2 a s

v_t² – v_o² = 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 (v_o) = 20 m/s

Acceleration (a) = 4 m/s²

Distance (s) = 150 meters

Wanted : Final speed (v_t)

Solution :

v_t² = v_o² + 2 a s

v_t² = 20² + 2 (4)(150)

v_t² = 400 + 1200

v_t² = 1600

v_t = 40 m/s

The initial velocity of object 2 :

Known :

Final speed (v_t) = 50 m/s

Acceleration (a) = 3 m/s²

Distance (s) = 150 meters

Wanted : Initial speed (v_o)

Solution :

v_t² = v_o² + 2 a s

v_t² – 2 a s = v_o²

50² – 2(3)(150) = v_o²

2500 – 900 = v_o²

1600 = v_o²

v_o = 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 (v_o) = 2 m/s

The final speed (v_t) = 22 m/s

Distance (s) = 120 meters

Wanted : Distance

Solution :

v_t² = v_o² + 2 a s

v_t² – v_o² = 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 (v_t) = 24 m/s

Distance (s) = 140 meters

Wanted : Initial speed (v_o)

Solution :

v_t² = v_o² + 2 a s

24² = v_o² + 2 (2)(140)

576 = v_o² + 560

576 – 560 = v_o²

16 = v_o²

v_o = 4 m/s

Distance of object 3 :

Known :

Initial speed (v_o) = 0 m/s

Final speed (v_t) = 20 m/s

Acceleration (a) = 2 m/s²

Wanted : Distance (s)

Solution :

v_t² = v_o² + 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