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Example Question - velocity calculation

Here are examples of questions we've helped users solve.

Calculating Velocity and Force in Formula 1 Racing

Pour résoudre la partie "Défi 1 : Relations de base", nous utilisons l'équation cinématique de base pour trouver la vitesse \( v \). La distance \( d \) est donnée et nous connaissons l'accélération \( a \) et le temps \( t \). \[ v = u + at \] Puisque la voiture part de l'arrêt, \( u = 0 \), donc \( v = at \). Nous substituons \( a = 5 \,m/s^2 \) et \( t = 4 \,s \): \[ v = 5 \times 4 = 20 \,m/s \] Pour la "Défi 2 : Application de la deuxième loi de Newton", la force \( F \) est la masse \( m \) multipliée par l'accélération \( a \). \[ F = ma \] Substituons \( m = 605 \,kg \) (poids de la voiture plus le conducteur) et \( a = 5 \,m/s^2 \): \[ F = 605 \times 5 = 3025 \,N \] Ceci complète les calculs pour les défis 1 et 2.

Solving Physics Projectile Problem

Claro, vamos a resolver paso a paso la pregunta del problema de física mostrado en la imagen: El problema dice: "Se lanza verticalmente hacia arriba y desde el suelo una pelota de 0,75 kg con una velocidad inicial de 30 m/s. Calcula: a) La altura máxima que alcanzará la pelota. b) Su velocidad cuando se encuentra a 15 m de altura." a) Para calcular la altura máxima que alcanza la pelota, usaremos la conservación de energía, donde la energía cinética inicial se convertirá en energía potencial en el punto más alto. La energía cinética (Ec) inicial la podemos calcular con la fórmula \( Ec = \frac{1}{2}mv^2 \) y la energía potencial (Ep) en el punto más alto con \( Ep = mgh \), donde m es la masa, v es la velocidad, g es la aceleración debido a la gravedad (9.8 m/s^2 en la Tierra), y h es la altura. \( Ec_i = Ep_{max} \) \( \frac{1}{2}mv_i^2 = mgh_{max} \) Podemos cancelar la masa m de ambos lados de la ecuación ya que es igual en ambos términos y despejar para h: \( \frac{1}{2}v_i^2 = gh_{max} \) \( h_{max} = \frac{v_i^2}{2g} \) Sustituimos \( v_i = 30 m/s \) y \( g = 9.8 m/s^2 \): \( h_{max} = \frac{(30 m/s)^2}{2 \cdot 9.8 m/s^2} \) \( h_{max} = \frac{900 m^2/s^2}{19.6 m/s^2} \) \( h_{max} ≈ 45.92 m \) Por lo tanto, la altura máxima que alcanzará la pelota es aproximadamente 45.92 metros. b) Para calcular la velocidad de la pelota a 15 m de altura, usamos la conservación de energía nuevamente, pero esta vez considerando la energía cinética y potencial a esa altura: \( Ec_i + Ep_i = Ec_{15m} + Ep_{15m} \) Ya que la pelota empieza desde el reposo en el suelo, \( Ep_i \) es 0, y \( Ec_{15m} \) es lo que queremos encontrar. Así que: \( \frac{1}{2}mv_i^2 = \frac{1}{2}mv_{15m}^2 + mgh_{15m} \) Cancelamos m nuevamente y despejamos para \( v_{15m} \): \( \frac{1}{2}v_i^2 = \frac{1}{2}v_{15m}^2 + gh_{15m} \) Movemos \( gh_{15m} \) al otro lado y multiplicamos por 2 ambos lados para eliminar la fracción: \( v_i^2 - 2gh_{15m} = v_{15m}^2 \) Ahora, tomamos la raíz cuadrada para despejar \( v_{15m} \): \( v_{15m} = \sqrt{v_i^2 - 2gh_{15m}} \) Sustituimos \( v_i = 30 m/s \), \( g = 9.8 m/s^2 \) y \( h_{15m} = 15 m \): \( v_{15m} = \sqrt{(30 m/s)^2 - 2 \cdot 9.8 m/s^2 \cdot 15 m} \) \( v_{15m} = \sqrt{900 m^2/s^2 - 294 m^2/s^2} \) \( v_{15m} = \sqrt{606 m^2/s^2} \) \( v_{15m} ≈ 24.62 m/s \) Por lo tanto, la velocidad de la pelota cuando se encuentra a 15 m de altura es aproximadamente 24.62 m/s.

Calculating Time for Ball's Velocity

Para resolver la pregunta, primero necesitamos entender la relación entre el tiempo y la velocidad mostrada en la tabla. La velocidad de la pelota es proporcional al tiempo y podemos observar que para cada segundo adicional, la velocidad aumenta en 9.8 m/s. Esto se debe a la aceleración de la gravedad cerca de la superficie de la tierra, que es aproximadamente 9.8 m/s². En 5 segundos, la pelota alcanza una velocidad de 49 m/s. Para encontrar cuántos segundos más tardaría la pelota en alcanzar una velocidad de 93.1 m/s, primero calculamos la diferencia entre 93.1 m/s y la velocidad alcanzada al último tiempo registrado (49 m/s): 93.1 m/s - 49 m/s = 44.1 m/s Ahora dividimos la diferencia de velocidad por la aceleración para encontrar el tiempo adicional necesario: 44.1 m/s ÷ 9.8 m/s² = 4.5 segundos (aproximadamente). Por lo tanto, tardaría aproximadamente 4.5 segundos adicionales para que la pelota alcanzase una velocidad de 93.1 m/s.

Insufficient Information for Calculating Ball Speed

La imagen muestra una pregunta de un test o examen que dice: 9. ¿Qué velocidad llevaba la pelota a los 6,5 s? a) 63,1 m/s b) 60,3 m/s c) 63,7 m/s d) 65,3 m/s Para resolver este problema, necesitamos más información relacionada con el movimiento de la pelota, como su posición, velocidad inicial, aceleración o alguna otra información relevante que nos permita hacer un cálculo. Con la información proporcionada en la imagen, no hay suficientes datos para calcular la velocidad de la pelota a los 6,5 segundos. Si hay más información en otra parte del examen o libro de texto que esté relacionada con este problema, por favor, proporcione esa información adicional para que pueda asistirte en la solución de este ejercicio.

Calculating Heights and Velocities in Vertical Projectile Motion

Per risolvere la domanda presente nell'immagine, occorre utilizzare le leggi del moto uniformemente accelerato (caduta dei gravi), assumendo che l'accelerazione dovuta alla gravità sia g = 9,81 m/s² e trascurando la resistenza dell'aria. La domanda posta nell'immagine dice: "Un uomo che sta sul tetto di un edificio lancia una palla verticalmente verso l'alto; con velocità 12,2 m/s e la palla tocca terra dopo 4,25 s. Quale è la quota più alta raggiunta dalla palla? Quanto è alto l'edificio? Con quale velocità la palla tocca terra?" Per calcolare la quota più alta raggiunta dalla palla, possiamo usare la seguente formula del moto uniformemente accelerato per un corpo lanciato verticalmente verso l'alto: h_max = v_0^2 / (2 * g) dove h_max è l'altezza massima, v_0 è la velocità iniziale di lancio e g è l'accelerazione di gravità. Sostituendo i valori noti: h_max = (12,2 m/s)^2 / (2 * 9,81 m/s²) = 149,44 m²/s² / 19,62 m/s² ≈ 7,62 m Quindi, la quota più alta raggiunta dalla palla è di circa 7,62 metri. Per trovare l'altezza dell'edificio, possiamo usare la formula del tempo di volo totale per un corpo lanciato verticalmente: t_tot = (v_0 / g) + √(2 * h / g) Risolvendo per h (altezza dell'edificio): h = (t_tot - v_0 / g)² * g / 2 Sappiamo che il tempo totale è di 4,25 s. Sostituendo i valori noti nella formula: h = [(4,25 s - 12,2 m/s / 9,81 m/s²)]² * 9,81 m/s² / 2 h = [(4,25 s - 1,243 s)]² * 9,81 m/s² / 2 h = (3,007 s)² * 9,81 m/s² / 2 h = 9,042 s² * 9,81 m/s² / 2 h ≈ 88,34 m L'altezza dell'edificio è quindi di circa 88,34 metri. Infine, per la velocità con cui la palla tocca terra, possiamo usare la formula della velocità finale nel moto uniformemente accelerato nel caso di caduta libera (considerando che la velocità sarà negativa quando tocca terra perché è diretta verso il basso): v = v_0 - g * t Il tempo di caduta dal punto più alto è dato da: t_caduta = t_tot - t_salita = 4,25 s - 1,243 s ≈ 3,007 s Sostituendo i valori per trovare v: v = 12,2 m/s - 9,81 m/s² * 3,007 s v ≈ 12,2 m/s - 29,49 m/s v ≈ -17,29 m/s La velocità con cui la palla tocca terra è di circa 17,29 m/s, diretta verso il basso (indicata qui come negativa per convenzione).

Vector Addition for Airplane Velocity Calculation

To solve this problem, you need to use vector addition to calculate the resultant velocity of the airplane relative to the ground, taking into account the wind's effect. The airplane has an initial velocity of 70 m/s, and the wind has a velocity of 18 m/s at a 120° angle to the airplane's direction. First, let's define the vectors: - The airplane's velocity vector, \( \vec{V}_a \), is 70 m/s in the direction of the positive x-axis. - The wind's velocity vector, \( \vec{V}_w \), is 18 m/s at a 120° angle from the airplane's direction (which means it is 30° above the negative x-axis, or equivalently, 150° from the positive x-axis). Now, decompose \( \vec{V}_w \) into x and y components: - \( V_{wx} = \vec{V}_w \cos(150°) = 18 \cos(150°) = -18 \cos(30°) \) (since cos(150°) = -cos(30°)) - \( V_{wy} = \vec{V}_w \sin(150°) = 18 \sin(150°) = 18 \sin(30°) \) (since sin(150°) = sin(30°)) Calculate these components: - \( V_{wx} = -18 \times \cos(30°) = -18 \times \sqrt{3}/2 \) - \( V_{wy} = 18 \times \sin(30°) = 18 \times 1/2 \) Now, add the wind's components to the airplane's velocity to find the resultant velocity vector, \( \vec{V}_r \): - \( V_{rx} = V_{ax} + V_{wx} = 70 + (-18 \times \sqrt{3}/2) \) - \( V_{ry} = V_{ay} + V_{wy} = 0 + 18 \times 1/2 \) Calculate the magnitude of the resultant velocity: - \( |\vec{V}_r| = \sqrt{V_{rx}^2 + V_{ry}^2} \) Lastly, you'll want to calculate the distance traveled over 6 minutes. You can convert the time to seconds (6 minutes * 60 seconds/minute = 360 seconds) and then use \( d = |\vec{V}_r| \times t \) to find the distance \( d \). Plug in the values and calculate the numerical results.

Solving for Acceleration Due to Gravity in Physics Problem

The image shows a physics problem that reads: "If a ball is dropped and attains a velocity of 29.4 m/s in 3.00 s, what is the acceleration due to gravity?" To solve for the acceleration due to gravity, we can use the kinematic equation that relates initial velocity (v_i), final velocity (v_f), acceleration (a), and time (t): \[ v_f = v_i + at \] For an object that is dropped, the initial velocity is zero (\( v_i = 0 \)), so the equation simplifies to: \[ v_f = at \] We can now solve for acceleration (a) using the final velocity (v_f) and time (t) provided: \[ a = \frac{v_f}{t} \] Plugging in the given values: \[ a = \frac{29.4 \text{ m/s}}{3.00 \text{ s}} \] \[ a = 9.8 \text{ m/s}^2 \] So, the acceleration due to gravity is \( 9.8 \text{ m/s}^2 \), which is consistent with the average acceleration due to gravity on Earth.

Vector Function Derivative: Velocity Calculation

The provided image contains a vector function labeled as number 4: \[ \mathbf{r}(t) = 4i + 2\cos(t)j + 3\sin(t)k \] To solve this question, you would typically need more context, such as finding the derivative of the vector function for velocity, the second derivative for acceleration, or evaluating at a specific point t. However, since the question isn't provided other than the expression for the vector function, I will show you how to find the derivative, which is a common operation on vector functions in calculus. The derivative of a vector function is found by differentiating each of the component functions with respect to t. Here's how you would find the derivative of the given vector function \(\mathbf{r}(t)\): \[ \frac{d}{dt}\mathbf{r}(t) = \frac{d}{dt}(4i) + \frac{d}{dt}(2\cos(t)j) + \frac{d}{dt}(3\sin(t)k) \] For each of the components: 1. The derivative of a constant (4i) is 0. 2. The derivative of \(2\cos(t)j\) with respect to t is \(-2\sin(t)j\). 3. The derivative of \(3\sin(t)k\) with respect to t is \(3\cos(t)k\). Putting it all together, we get: \[ \mathbf{r}'(t) = 0i - 2\sin(t)j + 3\cos(t)k \] So the derivative of the vector function \(\mathbf{r}(t) = 4i + 2\cos(t)j + 3\sin(t)k\) is: \[ \mathbf{r}'(t) = -2\sin(t)j + 3\cos(t)k \] This vector function represents the velocity of a particle moving along the path defined by \(\mathbf{r}(t)\) at any time t.

Estimating Train Distance with Graph

The image contains a graph showing the velocity of a train over time and two questions: a) Use four strips of equal width to estimate the distance the train traveled in the first 20 seconds. b) Is your answer to part a) an underestimate or an overestimate? To solve part a), we need to estimate the area under the graph from 0 to 20 seconds using the four strips. Since precise measurements of the strips are not possible from the image, I'll explain the procedure. The x-axis likely represents time in seconds, and the y-axis represents velocity in meters per second (m/s), although the units are not visible. Assuming standard units, to estimate the distance using four strips, we would divide the 20-second interval into four equal parts, each representing 5 seconds. For each 5-second interval, estimate the average velocity and then multiply by 5 to find the approximate distance traveled in each interval. This is essentially calculating the area of each rectangular strip since Distance = Velocity x Time. To get a better estimate, you might average the velocity at the start and the end of each 5-second interval to find the height of each rectangle (due to the graph's linearity, it will be the velocity at the middle of each interval): Let's denote the velocity at the start of each interval as V1, V2, V3, and V4, and the velocity at the end as V5, V6, V7, and V8 respectively. The approximate distance (D) traveled in 20 seconds using this method would be: D ≈ (V1 + V5) / 2 * 5 + (V2 + V6) / 2 * 5 + (V3 + V7) / 2 * 5 + (V4 + V8) / 2 * 5 For part b), since the graph shows a curve that is concave up (the slope is increasing), using four strips will likely result in an underestimate. This is because the rectangles will not cover the entire area under the curve - there will be small triangular areas between the top of each rectangle and the curve that are not being accounted for.