The motion of a jet plane just after landing on a runway is described by the a-t graph. Determin

The motion of a jet plane just after landing on a runway is...

Key Concepts

Integration and Area Under the Curve

Integration plays a central role in converting acceleration information into velocity and displacement profiles. The area under the acceleration-time graph corresponds to the change in velocity over a given time interval. Similarly, the area under the velocity-time graph represents the displacement. This method allows for the determination of key events in the motion, such as when the object comes to rest.

Determining the Time When an Object Stops

Determining when an object stops involves finding the time at which its velocity becomes zero. By integrating the acceleration to obtain the velocity function and applying initial conditions, one can set the velocity equal to zero and solve for time. This procedure is a common application in problems involving deceleration or braking and is critical for designing motion scenarios.

Relationship between Acceleration, Velocity, and Displacement

This concept involves understanding how acceleration, velocity, and displacement are related through calculus. In kinematics, the acceleration function is the derivative of velocity with respect to time, and velocity is the derivative of displacement with respect to time. Conversely, velocity can be obtained by integrating the acceleration function over time, and displacement is the integral of the velocity function. This principle allows one to graphically convert an acceleration-time graph into velocity-time and displacement-time graphs using initial conditions.

Transcript

00:01 Hi everyone, what we have is the motion of a jet plane whose acceleration with regard to time is described by the acceleration time graph and we know the initial position is zero and the initial velocity is 300 feet per second and what we're asked to solve for is the time at which the jet stops and also construct the velocity versus time and position versus time graphs so what we can do is actually take this acceleration versus time graph and solve for the velocity versus time function just by integrating it since their acceleration is given to us with respect to time.

00:39 So just by looking at the acceleration versus time graph, there are three distinct sections that we can use to solve for the velocity versus time.

00:48 So the initial section is from t equals zero to t equals ten seconds.

00:53 And as you can see the acceleration is just zero.

00:58 Therefore, our velocity with respect to time will simply be our initial velocity.

01:03 So for t, so basically, for t from zero to 10 seconds, our acceleration with our velocity with respect to time will just be 300 feet per second since there is no acceleration.

01:25 Next, we're going to look from 10 to 20 seconds.

01:29 We see that our acceleration is linear.

01:34 So from 10 to 20 seconds, we'll see that our acceleration is defined by this linear function, such that acceleration equals t minus 30 feet per second square.

01:50 Now, using the initial condition that our velocities 300 feet, per second at 10 seconds, we can say that the integral from 300 to v of dv is equal to the integral from 10 seconds to t, t minus 30 d t.

02:13 Now carrying this integral out on both sides gives us the expression for velocity as a function of time for this portion from t equals 10 to 20 seconds, that velocity is equal to 1 half t squared minus 30 t plus 550.

02:32 Next we can see that at t equals 20, the acceleration is constant at negative 10 meters per second squared.

02:42 So what we can do is we can say that from 4t greater than 20 seconds, that the integral from the velocity at 20 seconds, which we compute to be 150 feet per second by plugging this in by plugging in 10 seconds or 20 seconds into our equation for velocity that we found here we see that the velocity at 20 seconds is 150 feet per second so integrating this from 150 to v of dv is equal to 20 to t of negative 10 d t which we got from our graph we see that the velocity as a function of time for t greater than 20 seconds is negative 10 t plus 350 now we can directly use this to solve for the total time that it's going to take to bring this aircraft to zero velocity so what we can simply do is say that zero will equal negative 10 t prime plus 350 solving this equation for t prime gives us that t prime is equal to 35 seconds.

03:56 So from this we actually have all the information we need to construct velocity versus time graph.

04:01 However, recall that we're also asked to plot our position versus time graphs.

04:05 We don't have enough information for that yet.

04:08 What we need to do is take our velocity functions and integrate them to get our position functions.

04:14 So for our initial time from 0 to 10 seconds, what we can do is integrate our velocity functions and say that integral from 0 to s of ds will be equal to then grow from 0 to t of 300 d t.

04:31 This gives us that s is equal to 300 t in feet from 0 to 10 seconds.

04:38 So by plugging in 10 seconds here we'll see that add 10 seconds the position is 3 ,000 feet.

04:47 What we can do is we can use this piece of information to move to our next portion of acceleration from 10 to 20 seconds when our acceleration is linear.

04:56 What we can do is say that the integral from 3 ,000 feet to s of ds is equal to the integral from 10 seconds to t of 1 .5t squared minus 30 t plus 550 d t.

05:14 Computing this integral gives us that the position from 10 to 20 seconds is equal to 1 .6t cubed minus 15 t squared plus 550 t minus 1 ,167.

05:31 And by plugging in for t equal to 20 seconds, at 20 seconds, our position is going to be equal to 5 ,167 feet at t equals 20...

Please give Ace some feedback