Physics, A particle is released as part of an experiment. Its speed t seconds after release is g

Physics, A particle is released as part of an experiment....

Key Concepts

Velocity Function

A velocity function describes how the speed of a particle changes over time, often including direction. In physics, this function allows us to determine not only the instantaneous speed at any moment but also how the particle's motion evolves by showing when it accelerates, decelerates, or changes direction.

Definite Integration

Definite integration is a mathematical tool used in kinematics to calculate the net change in a quantity over an interval. By integrating the velocity function over a specific time period, one can obtain the displacement or the total distance traveled, depending on whether the absolute value of velocity is integrated.

Displacement versus Distance Traveled

Displacement represents the net change in position of a particle, which can be found by directly integrating the velocity function over time, considering signs. In contrast, the distance traveled is calculated by integrating the absolute value of velocity, accounting for all movement regardless of direction, especially when the velocity function may change sign.

Finding Critical Points for Motion Reversal

Identifying critical points, such as when the velocity function equals zero, is essential in determining when a particle changes its direction. This concept is critical for partitioning the time domain into intervals where the particle moves in a consistent direction, ensuring accurate computation of the total distance traveled by summing over absolute displacements in each interval.

Transcript

00:01 So we're given this equation which tells us the speed of our particle in meters per second after t seconds.

00:06 And in order to figure out the distance traveled in the first five seconds, we're just going to have to take the integral of this v of t equation from zero to five.

00:14 So the integral of negative 0 .5t squared plus 10t dt.

00:22 And so the way that we can do this is we can split up this integral.

00:25 So we can actually take this negative point five out as well.

00:29 So we have negative 0 .5, the integral from 0 to 5 of t squared, plus 10 times the integral from 0 to 5 of t.

00:40 And so now we can just do these two integrals pretty simply using power rules.

00:45 So this would be equal to negative 0 .5 times t to the third from 0 to 5 plus 10.

00:56 And this is sorry this is divided by three times t squared divided by two from zero to five and so t to the third divided by three from zero to five would be just 125 divided by three so this is equal to negative point five times 125 divided by three since at t is equal to zero this is equal to zero and then we're adding 10 times t squared in this case five squared is 25 divided by 2, and then when t is equal to zero, again, t squared divided by 2 is equal to 0.

01:34 So now we can simplify a little bit.

01:36 So this is equal to negative 0 .5 times 125 divided by 3.

01:42 That's going to be equal to negative 125 divided by 6.

01:46 And then we're adding 5 times 25, which is 125.

01:52 So now we can multiply 125.

01:56 By 6, which is equal to 750.

02:00 So we have negative 125 divided by 6 plus 750 divided by 6, which is then equal to 625 divided by 6.

02:11 So the distance traveled would be 625 divided by 6 meters in the first 5 seconds.

02:18 And now what we want to do is find how long it traveled, or the distance traveled in the second five seconds...