# Linear motion – problems and solutions

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.

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

4. An object travels at a constant 36 km/hour during 5 seconds, then accelerated 1 m/s^{2} during 10 seconds and then decelerated 2 m/s^{2} 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 (v_{o}) = velocity in motion 1 = 10 m/s

Acceleration (a) = 1 m/s^{2}

Time interval (t) = 10 seconds

__Motion 3 = constant deceleration__

Deceleration (a) = -2 m/s^{2}

Final velocity (v_{t}) = 0 m/s

Wanted : which graph shown object’s travels

Solution :

__The final velocity of motion 2 :__

v_{t }= v_{o} + a t

v_{t} = 10 + (1)(10) = 10 + 10 = 20 m/s

Final velocity of motion 2 (v_{t}) = initial velocity of motion 3 (v_{o}) = 20 m/s

__Time interval of motion 3 :__

v_{t }= v_{o} + 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.

5. A car travels at a constant 5 m/s in 10 seconds, then accelerated 1 m/s^{2} 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 (v_{o}) = velocity in motion 1 = 5 m/s

Acceleration (a) = 1 m/s^{2}

Time interval (t) = 5 seconds

Final velocity (v_{t}) = 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 = v_{o} t + ½ a t^{2} = (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 :

v_{t} = v_{o} + a t

v_{t }= 5 + (1)(5) = 5 + 5 = 10 m/s

Final velocity for motion 2 (v_{t}) = initial velocity for motion 3 (v_{o}) = 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 :

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

0^{2} = 10^{2} + 2 a (50)

0 = 100 + 100 a

100 = -100 a

a = – 100 / 100

a = – 1 m/s^{2}

Time interval of motion 3 :

v_{t} = v_{o} + 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.

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 :

v_{o} = 36 km/hour = 36,000 meters / 3600 s = 10 m/s

v_{t} = 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 :

v_{t} = v_{o} + a t

d = v_{o} t + ½ a t^{2}

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

Acceleration :

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

20^{2} = 10^{2} + 2 a (150)

400 = 100 + 300 a

400 – 100 = 300 a

300 = 300 a

a = 300 / 300

a = 1 m/s^{2}

Time interval :

v_{t} = v_{o} + a t

20 = 10 + (1) t

20 – 10 = t

t = 10 seconds

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 (v_{o}) = 20 m/s

Final speed (v_{t}) = 0 m/s

Time interval (t) = 5 seconds

__Wanted :__ distance (d)

__Solution :__

deceleration :

v_{t} = v_{o} + a t

0 = 20 + a 5

-20 = 5a

a = -20/5

a = -4 m/s^{2}

Distance traveled by car :

d = v_{o} t + 1/2 a t^{2}

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.