Linear motion – problems and solutions

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.

Linear motion – problems and solutions 1

Solution

A-B = motion at constant acceleration, B-C = motion at constant velocity, C-D = motion at constant deceleration.

Linear motion – problems and solutions 2

2. Graph of linear motion shown in the figure below. What is the distance traveled by an object from 0 second – 8 seconds.

Linear motion – problems and solutions 3

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.

SolutionLinear motion – problems and solutions 4

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/s2 during 10 seconds and then decelerated 2 m/s2 until rest. Which graph (v-t) shows the object’s travels.

Linear motion – problems and solutions 4

Linear motion – problems and solutions 6

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.

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Graph B shows the object’s travels.

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.

Linear motion – problems and solutions 7

Linear motion – problems and solutions 7

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.

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

7. A car travels at a constant 20 m/s. Car rest in 5 seconds after decelerated. What is the distance traveled by car.

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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.

Linear motion problems and solutions 11

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.

  1. Question: What distinguishes linear motion from other types of motion? Answer: Linear motion refers to the movement of an object along a straight path in a single direction. Unlike rotational or oscillatory motion, there’s no change in the direction of the moving object in linear motion.
  2. Question: How does distance differ from displacement in linear motion? Answer: Distance is the total length of the path taken by an object in motion, while displacement is the straight-line distance between the initial and final positions, along with the direction. Displacement is a vector quantity, whereas distance is scalar.
  3. Question: Why can speed and velocity have different values for the same motion? Answer: Speed is a scalar measure of how fast something is moving without considering its direction, while velocity is a vector quantity that includes both magnitude and direction. For linear motion, if there’s any change in direction, the average speed might be different from the magnitude of the average velocity.
  4. Question: What does it mean when we say an object has uniform linear motion? Answer: An object in uniform linear motion moves in a straight line at a constant velocity. This means its speed and direction remain unchanged over time.
  5. Question: How can an object have a constant speed but a changing velocity in linear motion? Answer: For an object in linear motion, the only way this can occur is if the direction of the motion changes. However, in true linear motion, the direction remains consistent. If the direction changes, it’s technically no longer purely linear motion.
  6. Question: What role does acceleration play in linear motion? Answer: Acceleration indicates a change in velocity over time. In linear motion, this could mean an increase or decrease in speed or a change in direction (if the motion is not purely linear). A non-zero acceleration indicates that the object’s velocity is changing.
  7. Question: How does negative acceleration, or deceleration, manifest in linear motion? Answer: In the context of linear motion, negative acceleration (often termed deceleration) refers to a decrease in speed along the same straight path. It’s an acceleration in the direction opposite to the motion.
  8. Question: What’s the relationship between net force and acceleration in linear motion according to Newton’s second law? Answer: According to Newton’s second law, the net force acting on an object is proportional to the acceleration it undergoes and acts in the same direction. Specifically, , where is the net force, is the object’s mass, and is its acceleration.
  9. Question: If a car is moving in a straight line and comes to a stop, can we say its final displacement is zero? Answer: No. Displacement refers to the straight-line distance between the initial and final positions with direction. If the car has moved from its initial position before stopping, the displacement is not zero; only its final velocity is.
  10. Question: How does the concept of inertia relate to linear motion? Answer: Inertia is an object’s resistance to a change in its state of motion. For an object in linear motion, inertia implies that the object will continue moving in a straight line at a constant velocity unless acted upon by an external force.

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