Example Question - position
Here are examples of questions we've helped users solve.
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>
CamTutor
In regards to math, we are professionals.
Get In Touch
Email: camtutor.ai@gmail.com