At a time t seconds after it is thrown up in the air, a...

Key Concepts

Average Velocity

Average velocity is defined as the total displacement divided by the total time elapsed. In the context of projectile motion, it provides a measure of the overall speed and direction over a specified time interval, regardless of the variations in instantaneous speeds during that interval.

Instantaneous Velocity

Instantaneous velocity is the rate of change of position with respect to time at a specific moment. It is found by taking the first derivative of the position function, and it gives a precise measure of the speed and direction of an object at a given instant.

Acceleration

Acceleration is the rate of change of velocity with respect to time. In kinematics, it is obtained by taking the derivative of the velocity function (or the second derivative of the position function), and it indicates how quickly the velocity is changing in magnitude or direction.

Derivatives in Kinematics

Derivatives are fundamental in calculus for describing rates of change, and in kinematics they connect concepts like position, velocity, and acceleration. The first derivative of a position function gives the velocity function, while the second derivative gives the acceleration function.

Quadratic Functions in Projectile Motion

Quadratic functions frequently arise in projectile motion due to the constant acceleration (like gravity). The vertex of a quadratic function represents the maximum or minimum point, which, in the case of upward and downward trajectories, corresponds to the peak height of a projectile. Solving the quadratic equation may also be used to determine the time when the projectile returns to its original position (time of flight).

Time of Flight

Time of flight refers to the total duration an object remains in the air during projectile motion. It is typically determined by solving the quadratic equation representing the vertical displacement for the time when the object’s height is zero, taking into account the initial conditions of the motion.

Transcript

00:01 So we're given this function f of t, which tells us the height after t seconds of our tomato.

00:06 And for part a, we're going to use this function to find the average velocity during the first two seconds.

00:12 So how we're going to do this is we're first going to find the derivative of this function or the velocity function of our tomato.

00:18 So f prime of t is equal to using the power rule.

00:23 We just take this exponent here and we bring it down and multiply it by 4 .9.

00:27 So this would be negative 9 .8 times t.

00:33 And then we do the same thing except here is t is only to the first.

00:37 So we bring it down and we minus one to the exponent.

00:40 So this is times t to the 0 or 1.

00:44 And as for our constant, all constants go to 0 when we are taking derivatives.

00:49 So now that we have our derivative, what we want to find is the derivative at 0, which is just equal to 25.

00:57 And we also want to find the derivative at 2 and this is equal to negative 9 .8 times 2 plus 25 which is equal to negative 19 .6 .6 plus 25 which is equal to 5 .4 so now that we have these two velocities we can just add them together and and then divide by 2 to find the average velocity over this interval.

01:33 So if we do that, we have 25 plus 5 .4 divided by 2, which is equal to 30 .4 divided by 2, which is equal to 15 .2.

01:46 So this is 15 .2 meters per second is the average velocity over the interval from 0 to 2.

01:55 And now for part b, we're going to find the exact, or the instantaneous velocity of our tomato at t is equal to two and the way that we can do that is by just using our velocity function or our f prime of t at t is equal to two so we actually already did this to find the average velocity and the answer was 5 .4 so our answer here f prime of 2 is equal to 5 .4 meters per second and now for part c, we're going to find the acceleration at t is equal to 2, and the acceleration function is just the second derivative of our height function.

02:42 So if we look at the derivative again of our height function, it was negative 9 .8t plus 25.

02:48 So if we use the power rule, we'll see that this negative 9 .8t just becomes negative 9 .8, and our constant goes to 0.

02:55 So this is equal to negative 9 .8.

02:59 So at f double prime of 2, this is still equal to negative 9 .8.

03:08 So we have negative 9 .8 meters per second squared as our acceleration at t is equal to 2.

03:20 And now for d, we're going to figure out how high the tomato actually goes, or the high point that our tomato goes and the way that we're going to do this is we're going to look at our f prime function again and we're going to set it equal to zero so we're going to say zero is equal to negative 9 .8 t plus 25 and usually the zeros would just be wouldn't be enough to figure out if this was a maximum or a minimum it would just be a critical point but since we know that we that our function is modeling a tomato as it's going up and then coming down in the air we know that we're going to have a maximum point at some point, and then we're going to have no other minimum values or maximum values on our, as our tomato is going up and down...

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