Example Question - acceleration
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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.
Kinematics and Dynamics of Car Braking and Free Falling Object
<p>1) To find the total distance the car travels during braking and the time it takes:</p> <p>a) The deceleration \( a \) can be calculated using the relation \( a = \frac{{v^2 - u^2}}{{2s}} \), where initial velocity \( u = 50 \) m/s (since the car is initially moving at 50 km/h, we convert this to m/s by multiplying by \( \frac{{1000}}{{3600}} \)), final velocity \( v = 0 \) m/s (since the car stops), and \( s \) is the distance. But since we are given a 10% negative gradient, the effective deceleration is the sum of the deceleration due to braking and the deceleration due to the slope: \( a = a_{braking} + a_{slope} \).</p> <p>b) The deceleration due to the slope \( a_{slope} \) is given by \( a_{slope} = g\sin(\theta) \), where \( g = 9.81 \) m/s\(^2\) is the acceleration due to gravity, and \( \sin(\theta) \) can be approximated by the slope percentage over 100, thus \( \sin(\theta) \approx \frac{{10}}{{100}} = 0.1 \).</p> <p>c) Hence, \( a_{slope} = 9.81 \times 0.1 = 0.981 \) m/s\(^2\), and using \( a_{braking} = -0.3g = -0.3 \times 9.81 \) m/s\(^2\), we find \( a = a_{braking} + a_{slope} = -2.943 + 0.981 = -1.962 \) m/s\(^2\).</p> <p>d) Now, we can calculate the distance \( s \) using \( s = \frac{{v^2 - u^2}}{{2a}} = \frac{{0^2 - (50/3.6)^2}}{{2 \times (-1.962)}} \).</p> <p>e) The time \( t \) it takes to stop is given by \( t = \frac{{v - u}}{{a}} = \frac{{0 - (50/3.6)}}{{-1.962}} \).</p> <p>2) To find the speed at which the body impacts the water:</p> <p>a) Use the kinetic energy at impact \( K.E. = \frac{1}{2}mv^2 \) equating it to the potential energy at the start \( P.E. = mgh \).</p> <p>b) Since \( K.E. = P.E. \), we have \( \frac{1}{2}mv^2 = mgh \). After canceling mass \( m \), the equation simplifies to \( v^2 = 2gh \).</p> <p>c) Plug in \( g = 9.81 \) m/s\(^2\) and \( h = 10 \) m to find \( v = \sqrt{2 \times 9.81 \times 10} \).</p> <p>The calculations from the above steps will give you the distance the car travels during braking and the time it takes to stop, as well as the speed at which the object hits the water.</p>
Analyzing Graphs to Find Acceleration
<p>The problem involves analyzing a velocity vs time graph to find acceleration. To find the acceleration from a velocity vs time graph, we calculate the slope of the line, since slope \(\frac{\Delta y}{\Delta x}\) in this context represents acceleration \(\frac{\Delta velocity}{\Delta time}\).</p> <p>From the graph labeled "velocity vs time," we can see that the line is straight, which indicates a constant acceleration. We need to pick two points from the graph to calculate the slope. We can use the points (2, 100) and (4, 300).</p> <p>Applying the slope formula:</p> \[ \text{Slope (acceleration)} = \frac{\Delta velocity}{\Delta time} = \frac{300 \text{ m/s} - 100 \text{ m/s}}{4 \text{ s} - 2 \text{ s}} = \frac{200 \text{ m/s}}{2 \text{ s}} = 100 \text{ m/s}^2 \] <p>Therefore, the acceleration of the object is \(100 \text{ m/s}^2\).</p>
Motion Graph Analysis
<p>The solution involves analyzing the given velocity vs. time graph to determine the acceleration. The slope of the velocity vs. time graph, which is a straight line, represents the acceleration.</p> <p>To find the slope (\(a\)) of the velocity vs. time graph, which is a straight line, we use the formula:</p> <p>\[ a = \frac{\Delta v}{\Delta t} \]</p> <p>Inspecting the graph, we can estimate the change in velocity (\(\Delta v\)) from 0 to approximately 3000 m/s and the change in time (\(\Delta t\)) from 0 to 5 s.</p> <p>\[ a \approx \frac{3000 \text{ m/s}}{5 \text{ s}} \]</p> <p>\[ a \approx 600 \text{ m/s}^2 \]</p> <p>The slope of the velocity vs. time graph is approximately 600 m/s², and this value represents the acceleration.</p>
Calculating the Acceleration of a Mass Due to an Increased Horizontal Force
<p>To calculate the acceleration of the mass, we will use Newton's second law of motion which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma).</p> <p>Given that F1 is increased to 20 N and the acceleration 'a' is to be calculated, we can rearrange the formula to find acceleration:</p> <p>a = \frac{F}{m}</p> <p>However, the mass 'm' is not provided in the question. The solution cannot be determined without knowing the mass of the object. Assuming the mass was given as 'm', the acceleration 'a' of the mass would be:</p> <p>a = \frac{20\ N}{m}</p> <p>Without the specific value of 'm', we cannot calculate the numerical value of the acceleration 'a', but this is the formula that would be used if the mass were known.</p>
Solving for Acceleration and Final Velocity
Para resolver el problema proporcionado en la imagen, primero vamos a interpretar la información que se nos da: Tenemos un cuerpo de masa \( m = 10 \) kg inicialmente en reposo, y se le aplica una fuerza constante de \( F = 25 \) N durante un intervalo de tiempo \( t = 5 \) s. a) Para encontrar la aceleración \( a \) que adquiere el cuerpo, usamos la segunda ley de Newton, que establece que la fuerza aplicada sobre un objeto es igual al producto de la masa del objeto por su aceleración (\( F = ma \)). Despejamos la aceleración: \[ a = \frac{F}{m} \] Sustituimos los valores dados: \[ a = \frac{25 \text{ N}}{10 \text{ kg}} \] \[ a = 2.5 \text{ m/s}^2 \] Por lo tanto, la aceleración que adquiere el cuerpo es de \( 2.5 \text{ m/s}^2 \). b) Para hallar la velocidad \( v \) que poseerá el cuerpo al cabo de los 5 segundos, utilizamos la definición de aceleración, que es el cambio de velocidad sobre el cambio de tiempo (\( a = \Delta v / \Delta t \)). Como el objeto parte del reposo, su velocidad inicial es \( v_0 = 0 \) m/s, y la aceleración es constante, entonces podemos usar la fórmula: \[ v = v_0 + a \cdot t \] Sustituimos \( v_0 \) con 0 y los valores correspondientes: \[ v = 0 + 2.5 \text{ m/s}^2 \cdot 5\text{ s} \] \[ v = 12.5 \text{ m/s} \] Por lo tanto, la velocidad que poseerá el cuerpo a los 5 segundos será de \( 12.5 \text{ m/s} \).
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