Example Question - distance
Here are examples of questions we've helped users solve.
Finding Distance in a Triangle
<p>Дано треугольник ABC со сторонами AB = 5 см, BC = 8 см, AC = 9 см.</p> <p>Сначала находим полупериметр треугольника: </p> <p>s = (AB + BC + AC)/2 = (5 + 8 + 9)/2 = 11 см.</p> <p>Находим радиус вписанной окружности r: </p> <p>r = A / s, где A — площадь треугольника. Используем формулу Герона для нахождения A: </p> <p>A = √(s(s - AB)(s - BC)(s - AC)) = √(11(11 - 5)(11 - 8)(11 - 9)) = √(11 * 6 * 3 * 2) = √(396) = 6√11.</p> <p>Следовательно, r = (6√11) / 11.</p> <p>Теперь находим расстояние от точки K до точки M биссектрисы BM. Сначала находим длину BM: </p> <p>BM = (AC * AB) / (AB + AC) = (9 * 5) / (5 + 9) = 45/14.</p> <p>Теперь, зная BM и KL = r, можем найти KM: </p> <p>KM = BM - r.</p>
Calculate Rate of Change in Distance with Respect to Time
<p>Let s(t) be the distance travelled as a function of time t.</p> <p>The rate of change of distance with respect to time is the derivative of s with respect to t, which is the velocity v(t) = \frac{ds}{dt}.</p> <p>For constant velocity, the rate of change of distance with respect to time is constant.</p> <p>The rate of change over the 80 km travelled is simply the constant velocity, which can be calculated as v = \frac{\Delta s}{\Delta t} where \Delta s = 80 \text{ km} and \Delta t = t_f - t_i, the time taken to travel the last 80 km.</p> <p>Without additional specific information about the time interval \Delta t or the actual function s(t) for the motion, we cannot compute a numerical value for the rate of change.</p>
Rate of Change in Distance Over Time for Vehicles
<p>Given that the train travels 150 km in 3 hours, the rate of change in distance with respect to time can be calculated as follows:</p> <p>The rate of change is the velocity of the train in km/min, which is constant since the question doesn't provide any information about the acceleration or deceleration of the train.</p> <p>First, convert the time from hours to minutes: \(3 \text{ hours} = 3 \times 60 \text{ minutes} = 180 \text{ minutes}.\)</p> <p>Then calculate the velocity of the train: \(\text{Velocity} = \frac{\text{Distance}}{\text{Time}} = \frac{150 \text{ km}}{180 \text{ minutes}} = \frac{5}{6} \text{ km/min}.\)</p> <p>The train travels at a constant velocity of \( \frac{5}{6} \text{ km/min}\) for the first 42 minutes.</p> <p>Thus, the rate of change of distance with respect to time, \(v\), for the first 42 minutes is \(\frac{5}{6} \text{ km/min}\).</p>
Vector Displacement and Distance Calculation
<p>Ilustramos el problema con un diagrama de vectores:</p> <p>Punto inicial O. El coche se mueve 52 km al Este, llegando al punto A. Luego se mueve 27 km al Sur, llegando al punto B.</p> <p>\[\vec{OA} = 52 \text{ km Este}, \quad \vec{AB} = 27 \text{ km Sur}\]</p> <p>La distancia es la longitud total del camino recorrido, es decir:</p> <p>\[ \text{Distancia} = |\vec{OA}| + |\vec{AB}| = 52 \text{ km} + 27 \text{ km} = 79 \text{ km} \]</p> <p>El desplazamiento es el vector resultante desde el punto de inicio al punto final, es decir \(\vec{OB}\). Calculamos su magnitud utilizando el teorema de Pitágoras, donde el desplazamiento es la hipotenusa de un triángulo rectángulo con lados 52 km y 27 km:</p> <p>\[ |\vec{OB}| = \sqrt{52^2 + 27^2} \]</p> <p>\[ |\vec{OB}| = \sqrt{2704 + 729} \]</p> <p>\[ |\vec{OB}| = \sqrt{3433} \]</p> <p>\[ |\vec{OB}| \approx 58.59 \text{ km} \]</p> <p>Por tanto, la distancia recorrida es de 79 km y el desplazamiento es aproximadamente de 58.59 km.</p>
Calculating Distance and Displacement
<p>La distancia es la longitud total del camino recorrido por el auto, que es igual a la suma de los desplazamientos en cada dirección:</p> \[ \text{Distancia} = 52 \text{ km} + 27 \text{ km} = 79 \text{ km} \] <p>El desplazamiento es el vector que va desde el punto inicial al final.</p> <p>Para calcular la magnitud del desplazamiento, usamos el teorema de Pitágoras, considerando un desplazamiento de 52 km hacia el Este y 27 km hacia el Sur, que forman un triángulo rectángulo.</p> \[ \text{Desplazamiento} = \sqrt{(52 \text{ km})^2 + (27 \text{ km})^2} \] \[ \text{Desplazamiento} = \sqrt{2704 + 729} \] \[ \text{Desplazamiento} = \sqrt{3433} \] \[ \text{Desplazamiento} \approx 58.6 \text{ km} \]
Analyzing the Relationship Between Variables in a Formula 1 Context
Le problème porte sur la relation entre la variation de vitesse (\(\Delta v\)), la variation de distance (\(\Delta x\)), et la variation de temps (\(\Delta t\)) pour une voiture de Formule 1. <p>Étape 1 : Déterminer la variation de vitesse en utilisant les données fournies. La variation de vitesse est la différence entre la vitesse finale et la vitesse initiale :</p> \[ \Delta v = v_f - v_i \] <p>Étape 2 : Trouver la variation de distance parcourue, qui est également donnée :</p> \[ \Delta x \] <p>Étape 3 : Calculer la variation de temps (\(\Delta t\)) à partir des informations fournies sur la distance et la vitesse :</p> \[ \Delta t = \frac{\Delta x}{v_m} \] où \( v_m \) est la vitesse moyenne pendant l'intervalle de temps considéré. <p>Étape 4 : Utiliser ces informations pour établir la relation demandée entre \(\Delta v\), \(\Delta x\), et \(\Delta t\). En physique, cette relation pourrait s'appuyer sur des équations cinématiques, mais les informations spécifiques nécessaires pour calculer la relation ne sont pas complètement visibles dans l'image fournie.</p> <p>Sans l'accès complet aux données et à la relation mathématique spécifique requise (comme l'équation de mouvement uniformément accéléré), il n'est pas possible de donner une solution détaillée spécifique. Cependant, en générale dans le contexte du mouvement uniformément accéléré, on pourrait appliquer l'équation suivante :</p> \[ v_f = v_i + a\Delta t \] où \( a \) est l'accélération. <p>En combinant cela avec l'équation de la distance pour un mouvement uniformément accéléré, on peut obtenir :</p> \[ \Delta x = v_i\Delta t + \frac{1}{2}a(\Delta t)^2 \] L'application correcte de ces équations nécessiterait des valeurs numériques pour \( v_i \), \( v_f \), et \( a \) qui ne sont pas fournies dans l'image.
Calculating Average Speed of Mary's Motorcycle Ride
<p>To find the average speed, we need to calculate the total distance traveled and the total time taken.</p> <p>Total distance = 35 km + 105 km = 140 km</p> <p>Total time = 0.5 hours (30 minutes) + 1.5 hours = 2 hours</p> <p>Average speed = \(\frac{\text{Total distance}}{\text{Total time}}\)</p> <p>Average speed = \(\frac{140 \text{ km}}{2 \text{ hours}} = 70 \text{ km/hour}\)</p>
Calculating Average Speed for Two Distinct Intervals of a Motorcycle Ride
<p>To find the average speed for each of the intervals, we will treat each interval separately, then find the total average speed.</p> <p>For the first interval:</p> <p>Speed = \frac{Distance}{Time} = \frac{35 \text{ km}}{0.5 \text{ hours}} = 70 \text{ km/h}</p> <p>For the second interval:</p> <p>Speed = \frac{105 \text{ km}}{1.5 \text{ hours}} = 70 \text{ km/h}</p> <p>Since the speed in both intervals is the same, the average speed for the entire trip is also 70 km/h.</p>
Relative Motion Problem Involving Two Trains
Dado que este es un problema de matemáticas, proporcionaré los pasos para llegar a la solución utilizando formato válido de LaTeX. <p>Definamos las siguientes variables:</p> <p>\( d \) = Distancia total entre A y B</p> <p>\( d_A \) = Espacio recorrido por el tren A</p> <p>\( d_B \) = Espacio recorrido por el tren B</p> <p>\( v_A \) = Velocidad del tren A</p> <p>\( v_B \) = Velocidad del tren B</p> <p>\( t \) = Tiempo hasta que se encuentren</p> <p>Sabemos que la distancia total entre A y B es \( d = 600 \, \text{km} \), la velocidad del tren A es \( v_A = 100 \, \text{km/h} \) y la velocidad del tren B es \( v_B = 80 \, \text{km/h} \).</p> <p>La distancia recorrida por ambos trenes hasta encontrarse será igual a la distancia total entre A y B:</p> \[ d_A + d_B = d \] \[ v_A \cdot t + v_B \cdot t = 600 \, \text{km} \] <p>Expresamos la distancia recorrida por cada tren en función del tiempo \( t \):</p> \[ 100t + 80t = 600 \] <p>Sumamos las distancias recorridas y resolvemos para \( t \):</p> \[ 180t = 600 \] \[ t = \frac{600}{180} \] \[ t = \frac{20}{6} \] \[ t = \frac{10}{3} \, \text{horas} \] <p>Ahora calculamos el espacio recorrido por cada tren al momento de encontrarse usando el tiempo \( t \):</p> \[ d_A = v_A \cdot t \] \[ d_A = 100 \cdot \frac{10}{3} \] \[ d_A = \frac{1000}{3} \, \text{km} \] \[ d_A = 333\frac{1}{3} \, \text{km} \] <p>Y de igual manera para el tren B:</p> \[ d_B = v_B \cdot t \] \[ d_B = 80 \cdot \frac{10}{3} \] \[ d_B = \frac{800}{3} \, \text{km} \] \[ d_B = 266\frac{2}{3} \, \text{km} \] <p>Por lo tanto, el tren A recorrerá \( 333\frac{1}{3} \) km y el tren B recorrerá \( 266\frac{2}{3} \) km antes de encontrarse.</p>
Relative Train Motion Problem
\[ \begin{align*} \text{1. Encuentra la distancia total que los trenes deben recorrer para encontrarse:} \\ d_{\text{total}} &= 600 \text{ km} \\ \text{2. Establece la velocidad relativa de ambos trenes:} \\ v_{\text{relativa}} &= v_{A} + v_{B} = 80 \text{ km/h} + 70 \text{ km/h} = 150 \text{ km/h} \\ \text{3. Encuentra el tiempo que tardan en encontrarse usando la fórmula de velocidad relativa:} \\ t &= \frac{d_{\text{total}}}{v_{\text{relativa}}} = \frac{600 \text{ km}}{150 \text{ km/h}} = 4 \text{ horas} \\ \text{4. Utiliza el tiempo hallado para determinar la distancia recorrida por el tren A:} \\ d_{A} &= v_{A} \cdot t = 80 \text{ km/h} \cdot 4 \text{ h} = 320 \text{ km} \\ \text{5. Ahora, encuentra la distancia separando A del punto de encuentro y llamémosla \( X \):} \\ X &= 600 \text{ km} - d_{A} = 600 \text{ km} - 320 \text{ km} = 280 \text{ km} \end{align*} \]
Calculating the Average Speed of a Student Walking Between Two Points
Given: - Distance from A to B (one direction) is 300 m. - Number of round trips in 20 minutes is 6. The total distance covered for 6 round trips (back and forth) is: \[ 6 \text{ round trips} \times 2 \times 300 \text{ m/round trip} = 3600 \text{ m} \] Time taken for 6 round trips is 20 minutes, converting minutes to seconds: \[ 20 \text{ minutes} \times 60 \text{ seconds/minute} = 1200 \text{ seconds} \] The average speed, \( V \), is total distance divided by total time: \[ V = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{3600 \text{ m}}{1200 \text{ s}} = 3 \text{ m/s} \]
Solving Distance and Average Speed Problem
Bài toán bạn cung cấp trong hình hỏi về quãng đường mà một người đi xe máy từ A đến B và ngược lại. Đây là bài toán về dạng toán chuyển động. Cụ thể, đây là vấn đề về quãng đường và vận tốc trung bình. Dựa trên thông tin trong hình, chúng ta có thể giải bài toán như sau: Ta có công thức tính vận tốc trung bình: V_tb = (S_đi + S_về) / T trong đó, S_đi là quãng đường từ A đến B, S_về là quãng đường từ B về A, và T là tổng thời gian cho cả hai hành trình. Biết rằng quãng đường từ A đến B và từ B về A là như nhau. Gọi S là quãng đường từ A đến B (hoặc từ B về A). Vậy tổng quãng đường: S_đi + S_về = S + S = 2S Thời gian để đi từ A đến B là 3 giờ 20 phút. Đổi 20 phút ra giờ ta được: 20 phút = 20/60 giờ = 1/3 giờ. Vậy tổng thời gian: T = 3 giờ + 1/3 giờ + 3 giờ = 6 1/3 giờ Đổi 6 1/3 giờ ra dạng giờ phù hợp để tính toán dễ dàng hơn: 6 1/3 giờ = 6 + 1/3 = 19/3 giờ Theo đề bài, vận tốc trung bình là 40km/h. Vậy ta có: V_tb = 2S / (19/3) = 40km/h Từ đó ta tìm được quãng đường S: 2S = V_tb * (19/3) 2S = 40 * (19/3) 2S = 760/3 S = (760/3) / 2 S = 760/6 S = 127km Vậy quãng đường từ A đến B là 127km.
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