Question

Тело двигалось прямолинейно 10 минут: сначала какое-то время равноускоренно, затем сразу равнозамедленно до полной остановки. Путь, пройденный телом, составил 3 км. Какова максимальная скорость тела во время движения? Ответ дайте в м/с.​

Answer 1

Путь - площадь под графиком скорости .

По условию - график скорости - треугольник от нулевых значений.

Площадь треугольника - половина произведения основания ( 600 секунд) на высоту ( максимальная скорость )

Откуда максимальная скорость равна два пути деленое на время

2*3000 / 600 = 10 м/с

Answer 2

Problem Analysis

We are given that an object moves in a straight line for 10 minutes. It starts with some time of constant acceleration and then immediately decelerates uniformly until it comes to a complete stop. The total distance covered by the object is 3 km. We need to find the maximum speed of the object during its motion in meters per second.

Solution

To solve this problem, we can break it down into two parts: the first part where the object is accelerating and the second part where it is decelerating.

Let's assume: - The time taken for acceleration is t1 minutes. - The time taken for deceleration is t2 minutes. - The maximum speed reached during acceleration is v_max m/s. - The maximum speed reached during deceleration is also v_max m/s.

We know that the total distance covered by the object is 3 km, which is equal to 3000 meters. We can use the equations of motion to find the values of t1, t2, and v_max.

Calculation

1. The distance covered during acceleration can be calculated using the equation: s1 = (1/2) * a * t1^2, where s1 is the distance covered during acceleration and a is the acceleration.

2. The distance covered during deceleration can be calculated using the equation: s2 = v_max * t2 - (1/2) * a * t2^2, where s2 is the distance covered during deceleration.

3. The total distance covered is the sum of s1 and s2: s1 + s2 = 3000.

4. The time taken for the entire motion is 10 minutes, which is equal to 600 seconds: t1 + t2 = 600.

5. The maximum speed reached during acceleration and deceleration is the same, which is v_max.

We have two equations and two unknowns (t1 and t2). We can solve these equations simultaneously to find the values of t1 and t2.

Solution Steps

1. Rearrange the equation s1 + s2 = 3000 to solve for t2: t2 = (3000 - s1) / v_max.

2. Substitute the value of t2 in the equation t1 + t2 = 600 to solve for t1: t1 + (3000 - s1) / v_max = 600.

3. Rearrange the equation to solve for t1: t1 = 600 - (3000 - s1) / v_max.

4. Substitute the values of s1 and v_max from the given information: - s1 = (1/2) * a * t1^2 - v_max = a * t1

5. Solve the equation s1 = (1/2) * a * t1^2 for a: a = (2 * s1) / t1^2.

6. Substitute the value of a in the equation v_max = a * t1 to solve for v_max: v_max = ((2 * s1) / t1^2) * t1.

7. Simplify the equation to find the value of v_max: v_max = (2 * s1) / t1.

8. Substitute the values of s1 and t1 from the given information: - s1 = (1/2) * a * t1^2 - t1 = 600 - (3000 - s1) / v_max.

9. Solve the equation to find the value of v_max.

10. Convert the value of v_max from km/h to m/s by multiplying it by 1000/3600.

Calculation

Let's calculate the value of v_max using the given information.

From the given information, we know that the total distance covered by the object is 3 km, which is equal to 3000 meters.

Let's assume a value for s1 (distance covered during acceleration) and calculate the corresponding values of t1 and v_max.

Assume s1 = 1000 meters.

Using the equation s1 = (1/2) * a * t1^2, we can solve for t1: 1000 = (1/2) * a * t1^2.

Simplifying the equation, we get: t1^2 = (2000 / a).

Using the equation t1 = 600 - (3000 - s1) / v_max, we can solve for v_max: v_max = (3000 - s1) / (600 - t1).

Substituting the value of t1^2 from the first equation, we get: v_max = (3000 - s1) / (600 - sqrt(2000 / a)).

Now, let's substitute the value of s1 = 1000 meters and calculate the value of v_max.

Using the equation v_max = (3000 - s1) / (600 - sqrt(2000 / a)), we get: v_max = (3000 - 1000) / (600 - sqrt(2000 / a)).

Simplifying the equation, we get: v_max = 2000 / (600 - sqrt(2000 / a)).

Now, let's calculate the value of v_max using the given information.

Answer

The maximum speed of the object during its motion is approximately v_max m/s.

Note: The exact value of v_max depends on the value of s1 (distance covered during acceleration) and a (acceleration), which are not provided in the given information. Therefore, we cannot provide an exact value for v_max without these additional details.