Question

Помогите решить задачу. Мальчик сьехал на санках с горы длиной 40м за 10сек., а потом проехал по горизонтальном участке еще 20м и остановился. Определить скорость в конце горы, ускорение на каждом участке.

Answer 1

Движение с горы - равноускоренное с ускорением а1. Найдем его:S= a1t^2/2,   a1 = (2S)/(t^2)Тогда скорость в конце горы:v = a1t = 2S/t = 8 м/с.Движение по горизонтальному участку - равнозамедленное, с ускорением -а2Уравнения кинематики:v-a2t2 = 0                                                         t2 = v/a2s= vt2 - (a2t2^2)/2   Из этой системы получим: s = v^2/2a2.Отсюда находим t2: t2 = 2s/v = 40/8 = 5 c.Общее время движения: t + t2 = 15 c.Средняя скорость по определению (весь путь поделить на все время):Vср = (S+s)/(t+t2) = 60 / 15 = 4м/с.Ответ: 8 м/с;  15 с;  4 м/с.

Answer 2

Calculating Speed and Acceleration of the Boy on the Sled

To calculate the speed at the end of the hill and the acceleration on each segment, we can use the basic kinematic equations of motion. Let's start by calculating the speed at the end of the hill and then determine the acceleration on each segment.

Speed at the End of the Hill: The speed at the end of the hill can be calculated using the formula: \[ v = u + at \] where: - \( v \) = final velocity - \( u \) = initial velocity - \( a \) = acceleration - \( t \) = time

Acceleration on Each Segment: The acceleration on each segment can be calculated using the formula: \[ a = \frac{{v - u}}{{t}} \] where: - \( a \) = acceleration - \( v \) = final velocity - \( u \) = initial velocity - \( t \) = time

Let's proceed with the calculations.

Speed at the End of the Hill:

The initial velocity (\( u \)) at the top of the hill is 0 m/s, as the boy starts from rest. The distance (\( s \)) covered on the hill is 40 m, and the time (\( t \)) taken is 10 seconds.

Using the formula \( v = u + at \), we can calculate the final velocity (\( v \)) at the end of the hill.

\[ v = u + at \] \[ v = 0 + a \times 10 \] \[ v = 10a \]

Acceleration on the Hill:

The acceleration (\( a \)) on the hill can be calculated using the formula \( a = \frac{{v - u}}{{t}} \).

\[ a = \frac{{v - u}}{{t}} \] \[ a = \frac{{10a - 0}}{{10}} \] \[ a = a \]

This means that the acceleration on the hill is equal to \( a \).

Speed on the Horizontal Segment:

After the hill, the boy travels 20 m on a horizontal segment. Since there is no mention of additional forces or resistance, we can assume that the speed remains constant on this segment.

Conclusion:

The speed at the end of the hill is \( 10a \) m/s, and the acceleration on the hill is \( a \). On the horizontal segment, the speed remains constant.

Let's solve for the values of speed and acceleration.