การเคลื่อนที่เชิงเส้น – ปัญหาและวิธีแก้ไข

การเคลื่อนที่เชิงเส้น – ปัญหาและวิธีแก้ไข

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 ความเร่งคงที่, B-C = motion at ความเร็วคงที่, C-D = motion at constant deceleration.

2. Graph of linear motion shown in the figure below. What is the ระยะทาง 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

พื้นที่ 1 = พื้นที่ของสามเหลี่ยม = ½ (2-0)(4-0) = ½ (2)(4) = 4

พื้นที่ 2 = พื้นที่ของสี่เหลี่ยมผืนผ้า = (6-2)(4-0) = (4)(4) = 16

พื้นที่ 3 = พื้นที่ของสามเหลี่ยม = ½ (8-6)(4-0) = ½ (2)(4) = 4

พื้นที่ 4 = พื้นที่ของสามเหลี่ยม = 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.

เป็นที่รู้จัก :

Motion 1 = constant velocity

Constant velocity (v) = 36 km/jam = 36 (1000 m) / 3600 s = 36,000 m / 3600 s = 10 m/s

ช่วงเวลา (t) = 5 วินาที

Motion 2 = constant acceleration

ความเร็วเริ่มต้น (v)_o) = velocity in motion 1 = 10 m/s

ความเร่ง (a) = 1 m/ ^s²

ช่วงเวลา (t) = 10 วินาที

Motion 3 = constant deceleration

Deceleration (a) = -2 m/s²

ความเร็วสุดท้าย (v _t ) = 0 ม./วินาที

Wanted : which graph shown object’s travels

วิธีการแก้ปัญหา:

The final velocity of motion 2 :

v _t = v _o + at

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 ม./วินาที

Time interval of motion 3 :

v _t = v _o + at

0 = 20 + (-2)(t)

0 = 20 – 2t

20 = 2t

เสื้อ = 20 / 2

t = 10 วินาที

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/s² in 5 seconds. Then decelerated until rest after car travels 137.5 meters. Which graph (v-t) shows the object’s travels.

เป็นที่รู้จัก :

Motion 1 = motion at constant velocity

Constant velocity (v) = 5 m/s

ช่วงเวลา (t) = 10 วินาที

Motion 2 = motion at constant acceleration

ความเร็วเริ่มต้น (v)_o) = velocity in motion 1 = 5 m/s

ความเร่ง (a) = 1 m/ ^s²

ช่วงเวลา (t) = 5 วินาที

ความเร็วสุดท้าย (v _t ) = 0 ม./วินาที

Total distance (s) = 137.5 m/s

Wanted : Graph shows car’s travel

วิธีการแก้ปัญหา:

Distance for motion 1 :

s = vt = (5)(10) = 50 เมตร

Distance for motion 2 :

ส = ว_o t + ½ ที่² = (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 + at

v_t = 5 + (1)(5) = 5 + 5 = 10 ม./วินาที

Final velocity for motion 2 (v_t) = initial velocity for motion 3 (v_o) = 10 ม./วินาที

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 ad

0 ² = 10 ² + 2 a (50)

0 = 100 + 100 a

100 = -100 ก

a = – 100 / 100

a = – 1 ม./วินาที^²

Time interval of motion 3 :

v _t = v _o + at

0 = 10 + (-1) t

0 = 10 – t

t = 10 วินาที

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.

เป็นที่รู้จัก :

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 เมตร

Wanted : time interval (t)

วิธีการแก้ปัญหา:

Three equation of motion at constant acceleration :

v _t = v _o + at

d = v _o t + ½ at ²

v _t ² = v _o ² + 2 ad

Acceleration :

v _t ² = v _o ² + 2 ad

20 ² = 10 ² + 2 a (150)

400 = 100 + 300 a

400 – 100 = 300 ก.

300 = 300 ก.

a = 300 / 300

a = 1 ม./วินาที^²

ช่วงเวลา :

v _t = v _o + at

20 = 10 + (1) t

20 – 10 = t

t = 10 วินาที

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

เป็นที่รู้จัก :

ความเร็วเริ่มต้น (v _o ) = 20 ม./วินาที

ความเร็วสุดท้าย (v _t ) = 0 ม./วินาที

ช่วงเวลา (t) = 5 วินาที

เป็นที่ต้องการ : ระยะทาง (ด)

วิธีการแก้ปัญหา:

deceleration :

v _t = v _o + at

0 = 20 + a 5

-20 = 5a

a = -20/5

a = -4 ม./วินาที^²

Distance traveled by car :

d = v_o t + 1/2 ที่²

d = (20)(5) + 1/2 (-4)(5)² = (20)(5) + (-2)(25)

d = 100 – 50

d = 50 เมตร

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

วัตถุที่ 1 :

วี = ส / ที

วี = ความเร็ว, s = ระยะทาง, t = ช่วงเวลา

v = 60/15 = 4 ซม./วินาที

v = 80/20 = 4 ซม./วินาที

v = 100/25 = 4 ซม./วินาที

Constant velocity = uniform linear motion.

วัตถุที่ 2 :

วี = ส / ที

วี = velocity, s = ระยะทาง, t = ช่วงเวลา

v = 4/2 = 2 ซม./วินาที

v = 9/3 = 3 ซม./วินาที

v = 25/5 = 5 ซม./วินาที

Velocity increases = nonuniform linear motion, accelerated.

1. คำถาม: What distinguishes linear motion from other types of motion? คำตอบ: 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. คำถาม: How does distance differ from displacement in linear motion? คำตอบ: 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. คำถาม: Why can speed and velocity have different values for the same motion? คำตอบ: 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. คำถาม: What does it mean when we say an object has uniform linear motion? คำตอบ: 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. คำถาม: How can an object have a constant speed but a changing velocity in linear motion? คำตอบ: 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. คำถาม: What role does acceleration play in linear motion? คำตอบ: 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. คำถาม: How does negative acceleration, or deceleration, manifest in linear motion? คำตอบ: 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. คำถาม: What’s the relationship between net force and acceleration in linear motion according to Newton’s second law? คำตอบ: 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, $F=ma$ที่นี่มี $F$ is the net force, $m$ is the object’s mass, and $a$ is its acceleration.
9. คำถาม: If a car is moving in a straight line and comes to a stop, can we say its final displacement is zero? คำตอบ: 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. คำถาม: How does the concept of inertia relate to linear motion? คำตอบ: 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.