# 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

**Distance traveled by object 3 ****:**

__Known ____:__

Initial speed (v_{o}) = 0 m/s

Final speed (v_{t}) = 30 m/s

Acceleration (a) = 3 m/s^{2}

__Wanted ____:__ Distance

__Solution :__

v_{t}^{2} = v_{o}^{2} + 2 a s

v_{t}^{2} – v_{o}^{2} = 2 a s

30^{2} – 0^{2 }= 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/s^{2}

Distance (s) = 150 meters

__Wanted :__ Final speed (v_{t})

__Solution :__

v_{t}^{2} = v_{o}^{2} + 2 a s

v_{t}^{2} = 20^{2} + 2 (4)(150)

v_{t}^{2} = 400 + 1200

v_{t}^{2} = 1600

**v**_{t }**= 40 m/s**

**The initial velocity of object 2 :**

__Known :__

Final speed (v_{t}) = 50 m/s

Acceleration (a) = 3 m/s^{2}

Distance (s) = 150 meters

__Wanted : Initial speed __(v_{o})

__Solution :__

v_{t}^{2} = v_{o}^{2} + 2 a s

v_{t}^{2} – 2 a s = v_{o}^{2}

50^{2} – 2(3)(150) = v_{o}^{2}

2500 – 900 = v_{o}^{2}

1600 = v_{o}^{2}

**v**_{o}** = 40 m/s **

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_{t}^{2} = v_{o}^{2} + 2 a s

v_{t}^{2} – v_{o}^{2} = 2 a s

22^{2} – 2^{2 }= 2 a (120)

484 – 4 = 240 a

480 = 240 a

a = 480/240

a = 2 m/s^{2}

**The initial speed of object 2 ****:**

__Known :__

Acceleration (a) = 2 m/s^{2}

Final speed (v_{t}) = 24 m/s

Distance (s) = 140 meters

__Wanted :__ Initial speed (v_{o})

__Solution :__

v_{t}^{2} = v_{o}^{2} + 2 a s

24^{2} = v_{o}^{2} + 2 (2)(140)

576 = v_{o}^{2} + 560

576 – 560 = v_{o}^{2}

16 = v_{o}^{2}

**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/s^{2}

__Wanted ____:__ Distance (s)

__Solution :__

v_{t}^{2} = v_{o}^{2} + 2 a s

20^{2 }= 0^{2} + 2 (2) s

20^{2 }= 2 (2) s

400 = 4 s

s = 400/4

**s = 100 meter****s**

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