# Bài tập đi lại thẳng biến đổi đều và Cách giải – Vật lý 10 chuyên đề

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Uniformly Variable Straight Walk Exercise and Solution. If in the previous lesson, the children had a good understanding of straight walking, then with the content and exercises on changing straight walking, it would not be difficult for them.

So, what are the types of exercises on straight walking? How to solve the problems of straight walking? Let’s find out through this article.

* Some formulas applied to solve uniformly variable straight walking exercises:

Formula for calculating acceleration:

– Formula for calculating instantaneous velocity vector: Formula for calculating distance traveled: – Relation formula between instantaneous velocity vector, acceleration and distance traveled (time-independent formula): * In there:

• If you move faster and faster, the acceleration: a > 0

• If you move slowly, acceleration: a<0

* uniformly variable forms of straight walking exercises:

° Form 1: Determine the vectors of instantaneous velocity, acceleration, and distance traveled in a straight line traveling uniformly

* Example 1: A car is running straight at a constant speed of 40 km/h, suddenly increasing the throttle to travel faster and faster. Calculate the acceleration of the car, knowing that after traveling a distance of 1 km, the car reaches a speed of 60 km/h.

See the solution

• Problem: A car is running straight at a constant speed of 40 km/h, suddenly increasing the throttle to travel faster and faster. Calculate the acceleration of the car, knowing that after traveling a distance of 1 km, the car reaches a speed of 60 km/h.

° Solution:

a) We have:

vo = 40 (km/h) = 40000(m)/3600(s) = 100/9 (m/s);

s = 1 (km) = 1000 (m);

v = 60 (km/h) = 60000(m)/3600(s) = 50/3 (m/s)

– apply the formula relating acceleration, instantaneous velocity vector and distance.  * Example 2: A train is traveling with an instantaneous speed vector v0 = 72 km/h, when the brakes are applied, it slows down steadily, after 10 seconds reaching the instantaneous speed vector v1 = 54 km/h

a) How long after braking does the train reach the instantaneous velocity vector v = 36 km/h and how long does it take the train to come to a complete stop?

b) Calculate the distance traveled by the train until it stops.

See the solution

• Problem: A train is traveling with an instantaneous velocity vector v0 = 72 km/h, when the brakes are applied to slow down, after 10 seconds, the instantaneous speed vector v1 = 54 km/h is reached.

a) How long after braking does the train reach the instantaneous velocity vector v = 36 km/h and how long does it take the train to come to a complete stop?

b) Calculate the distance traveled by the train until it stops.

– select the positive direction in the same direction as the train, the time origin is when the train starts to brake.

¤ Explanation:

a) We have: 72 km/h = 20 m/s; 54 km/h = 15 m/s

The acceleration of the train is: – The object reaches the instantaneous velocity vector v = 36km/h = 10m/s after time is:

We have:  – When the train comes to a complete stop, the object has an instantaneous velocity vector v’=0.  → After a period of 20s, the train from the instantaneous speed of 72(km/h) decreases to 36(km/h), and after 40s the train stops.

b) Applying the relation between the instantaneous velocity vector, acceleration and distance traveled, we have:  → So the train travels a distance of 400(m) and then stops.

° Form 2: Write uniformly variable linear equations

• Step 1: select a reference system

– Ox coordinate axis coincides with travel trajectory

– Origin (usually associated with the original position of the object)

– Time origin (usually when the object starts moving)

– Positive direction (usually chosen is the direction of travel of the object selected as a landmark)

• Step 2: From the selected frame of reference, determine the factors x0; v0; t0 of the object

> Note: v0 needs to specify the sign in the direction of travel

• Step 3: Write the travel equation

– The transformed linear equation of travel has the form: > Note:

– In this case, we need to consider the sign of travel, so we have: when the object moves faster and faster when the moving object slows down steadily

* The problem of finding the position, the moment when two objects meet:

Write down the equation of motion for each object

When two objects meet * Example: At 8 o’clock two objects travel in opposite directions on a distance AB 560m long. At A an object slows down uniformly with an acceleration of 0.2 m/s2. At B, the second body moves with uniform acceleration with an acceleration of 0.4 m/s2. We know that at A object one has an initial instantaneous velocity vector of 10 m/s, at B object two starts to move from a stationary position.

a) Write down the equations of motion for the two bodies

b) Determine when and where the two cars meet.

See the solution

• Problem: At 8 o’clock two objects travel in opposite directions on a distance AB of 560m long. At A an object slows down uniformly with an acceleration of 0.2 m/s2. At B, the second body moves with uniform acceleration with an acceleration of 0.4 m/s2. We know that at A object one has an initial instantaneous velocity vector of 10 m/s, at B object two starts to move from a stationary position.

a) Write down the equations of motion for the two bodies

b) Determine when and where the two cars meet.

¤ Explanation:

choose the origin at A, the time origin is at 8 o’clock, the positive direction is the direction from A to B.

a) The equations of motion of the two bodies are:

• Object 1: (first)

(object one slows down, so a, v have opposite signs; v > 0 ⇒ a < 0)

• Object 2: (2)

(object 2 moves faster and faster, so a and v have the same sign; opposite direction is positive, so v < 0 ⇒ a<0)

b) When the two cars meet, we have:  With t = 40 (receiving); t = -140 (type).

– Substituting t = 40(s) into equation (1) we get: → So 2 cars meet after 40(s) time at a position 240m from origin A.

° Form 3: Calculate the distance traveled by the object in the nth second and in the last n seconds

1. Distance traveled in nth second

Calculate the distance traveled by the object in n seconds: + Calculate the distance traveled by the object in (n-1) seconds: + Distance traveled in the nth second: 2. Distance traveled in last n seconds

Calculate the distance traveled in t seconds: + Calculate the distance traveled by the object in (t – n) seconds: + Distance traveled in last n seconds: * Example 1 (Lesson 14, page 22 of Physics Textbook 10): A train is running at 40 km/h when it brakes, walking straight and slowly slowing down to enter the station. After 2 minutes, the train stopped at the platform.

a) Calculate the acceleration of the train.

b) Calculate the distance traveled by the train while braking.

See the solution

• Problem: A train is running at a speed of 40 km/h when it brakes and slows down slowly to enter the station. After 2 minutes, the train stopped at the platform.

a) Calculate the acceleration of the train.

b) Calculate the distance traveled by the train while braking.

° Solution:

◊ We have (subject to):

– Initial: v0 = 40 (km/h) = 100/9 (m/s).

– After 2 minutes, ie t = 2 minutes = 120 s, the train stops so: v = 0

a) The acceleration of the train is: b) The distance traveled by the train while braking is:  → The distance traveled during braking is 666.7(m).

* Example 2 (Lesson 15, page 22, Physics Textbook 10): A motorbike is traveling at a speed of 36 km/h, and suddenly the driver sees a hole in front of him, 20 m away from the vehicle. He braked hard and the car came close to the mouth of the hole, then stopped.

a) Calculate the acceleration of the car.

b) Calculate the braking time.

See the solution

• Problem: A motorbike is traveling at a speed of 36 km/h, and suddenly the driver sees a hole in front of him, 20 m away from the vehicle. He braked hard and the car came close to the mouth of the hole, then stopped.

a) Calculate the acceleration of the car.

b) Calculate the braking time.

° Solution:

◊ We have (problem for):

– Initial: v0 = 36 (km/h) = 10 (m/s).

– Then the car brakes, after the distance S = 20 m the car stops: v = 0

a) The acceleration of the car is:  b) Braking time is: → Braking time is 4(s)

* Example 3: A car travels in a straight line at a constant speed with v0 = 10.8km/h. In the sixth second the car travels a distance of 14m.

a) Calculate the acceleration of the car

b) Calculate the distance traveled by the car in the first 10 seconds.

See the solution

• Problem: A car travels straight and steadily with v0 = 10.8km/h. In the sixth second the car travels a distance of 14m.

a) Calculate the acceleration of the car

b) Calculate the distance traveled by the car in the first 10 seconds.

¤ Explanation:

a) We have: 10.8 km/h = 3 m/s

The distance traveled by the object in 6 seconds is:  The distance traveled by the object in the first 5 seconds is:  – The distance traveled by the object in the sixth second is:  ⇒ The object’s acceleration at the sixth second is: b) The distance traveled by the object in the first 10 seconds is: → In the first 10s the object travels a distance of 130(m).

As you have seen, if you have understood and applied it well to solve the exercises in the straight walking section, then it is not difficult to solve the exercises in the content of constant straight walking. none.

Hopefully, with the above-mentioned training in solving skills of straight and variable walking exercises, it has helped them better understand the theoretical content, taking that as a basis to absorb the next lessons well, wish them all the best. Good.

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