Example Question - time
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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>
Variation in Velocity and Time in Formula 1 Racing
D'après le document, la formule pour la relation entre la force, la masse et l'accélération est \( F = ma \). La variation de la vitesse est \( \Delta v \), et la variation du temps est \( \Delta t \). On a: <p>\( F = \frac{m \Delta v}{\Delta t} \)</p> On peut réarranger cette formule pour résoudre le temps \( \Delta t \): <p>\( \Delta t = \frac{m \Delta v}{F} \)</p> Pour calculer le temps mis par les voitures de Formule 1 pour atteindre la vitesse de 100 km/h à partir du repos, on utilise: <p>\( \Delta v = v - u \)</p> \( v = 100 \) km/h (la vitesse finale) et \( u = 0 \) km/h (la vitesse initiale car la voiture part du repos). Il faut convertir la vitesse de km/h en m/s pour être cohérent avec l'unité de force (N) qui est en mètres par seconde carrée (m/s\(^2\)): <p>\( 100 \) km/h = \( \frac{100 \times 1000}{3600} \) m/s = \( \frac{1000}{36} \) m/s = \( \frac{250}{9} \) m/s</p> <p>\( \Delta v = \frac{250}{9} \) m/s</p> Maintenant, en insérant \( \Delta v \) et les valeurs pour \( m \) (masse de la voiture plus le pilote, en kg) et \( F \) (la force, en N) dans la formule du temps, on peut calculer \( \Delta t \): <p>\( \Delta t = \frac{m \times \frac{250}{9}}{F} \)</p> En utilisant les valeurs de masse et de force données pour Nico Hülkenberg (masse totale \( m = 661 \) kg et force \( F = 20000 \) N), on obtient: <p>\( \Delta t = \frac{661 \times \frac{250}{9}}{20000} \) s</p> <p>\( \Delta t = \frac{661 \times 250}{9 \times 20000} \)</p> <p>\( \Delta t = \frac{165250}{180000} \)</p> <p>\( \Delta t \approx 0.918 \) s</p> Et pour Carlos Sainz Jr (masse totale \( m = 654 \) kg et force \( F = 20000 \) N), on obtient: <p>\( \Delta t = \frac{654 \times \frac{250}{9}}{20000} \) s</p> <p>\( \Delta t = \frac{654 \times 250}{9 \times 20000} \)</p> <p>\( \Delta t = \frac{163500}{180000} \)</p> <p>\( \Delta t \approx 0.909 \) s</p> Ainsi, selon les données, Nico Hülkenberg prend environ 0.918 secondes et Carlos Sainz Jr prend environ 0.909 secondes pour atteindre la vitesse de 100 km/h à partir du repos, en supposant que ces accélérations sont constantes et que la force fournie est de 20000 N pour chaque voiture.
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 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 Distance Traveled by an Accelerating Car from Rest
<p>The distance traveled by an object under constant acceleration can be found using the equation:</p> <p>\[ s = ut + \frac{1}{2}at^2 \]</p> <p>where:</p> <p>\[ u \] is the initial velocity (in this case, 0 m/s, since the car starts from rest),</p> <p>\[ a \] is the constant acceleration (3 m/s^2), and</p> <p>\[ t \] is the time (4 seconds).</p> <p>Plugging in the values, we get:</p> <p>\[ s = (0) \cdot 4 + \frac{1}{2} \cdot 3 \cdot (4)^2 \]</p> <p>\[ s = 0 + \frac{1}{2} \cdot 3 \cdot 16 \]</p> <p>\[ s = 0 + \frac{1}{2} \cdot 48 \]</p> <p>\[ s = 24 \text{ meters} \]</p> <p>Thus, the distance traveled by the car is 24 meters.</p>
Calculating Final Velocity of a Bicycle Under Constant Acceleration
<p>v_{final} = v_{initial} + a * t</p> <p>v_{final} = 5 m/s + (2 m/s^2 * 3 s)</p> <p>v_{final} = 5 m/s + 6 m/s</p> <p>v_{final} = 11 m/s</p>
Calculating Final Velocity of a Car After Uniform Acceleration
<p>$v = u + at$</p> <p>Where:</p> <p>$v$ is the final velocity,</p> <p>$u$ is the initial velocity (which is $0$ m/s since the car starts from rest),</p> <p>$a$ is the acceleration ($2 \ m/s^2$),</p> <p>$t$ is the time ($5$ seconds).</p> <p>Substitute the given values into the equation:</p> <p>$v = 0 + (2 \ m/s^2)(5 \ s)$</p> <p>$v = 10 \ m/s$</p> <p>Therefore, the car's velocity after $5$ seconds is $10 \ m/s$.</p>
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: \[ V = \frac{\text{Total Distance}}{\text{Total Time}} = \frac{3600 \text{ m}}{1200 \text{ s}} = 3 \text{ m/s} \]
Distance Calculation for Motorcycle Route
Đề bài trong hình đưa ra cho ta biết rằng một chiếc xe máy xuất phát từ A lúc 7 giờ sáng và đi đến B với vận tốc không đổi là 30 km/h. Xe dừng lại ở B một thời gian sau đó quay lại A. Trên đường về, xe giảm vận tốc đi 5 km/h so với lúc đi. Biết rằng xe máy đến A lúc 11 giờ sáng cùng ngày đó, chúng ta cần tính quãng đường AB. Để giải bài toán này, ta cần tìm thời gian di chuyển của xe máy từ A đến B và từ B trở lại A. Gọi \( t \) (giờ) là thời gian xe máy đi từ A đến B. Do vận tốc xe máy lúc đi là 30 km/h, quãng đường AB có thể được tính bằng \( 30t \). Khi quay trở lại A từ B, xe máy có vận tốc là \( 30 - 5 = 25 \) km/h. Do thời gian đi từ A đến B và trở lại A là từ 7 giờ đến 11 giờ, tổng thời gian là 4 giờ, trừ đi thời gian xe máy đi từ A đến B, ta có thời gian trở lại là \( 4 - t \) giờ. Thời gian để quay trở lại A từ B với vận tốc 25 km/h là \( 25(4 - t) \), và quãng đường này cũng chính là AB, nên ta có: \[ 30t = 25(4 - t) \] Mở rộng phương trình: \[ 30t = 100 - 25t \] Cộng \( 25t \) vào cả hai vế và trừ 100 cho cả hai vế, ta có: \[ 30t + 25t = 100 \] Giải phương trình này cho \( t \): \[ 55t = 100 \] \[ t = \frac{100}{55} \] \[ t = \frac{20}{11} \] Tính giá trị của \( t \): \[ t \approx 1.82 \] giờ (với độ chính xác đến hai chữ số thập phân). Bây giờ ta có thể tình quãng đường AB: \[ AB = 30t \approx 30 \times 1.82 \approx 54.55 \] km. Vậy quãng đường AB khoảng 54.55 km.
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