Acceleration A drag racer accelerates at a(t)=88 ft / s^2 ....

Acceleration A drag racer accelerates at $a(t)=88 \mathrm{ft} / \mathrm{s}^{2} .$ Assume that $v(0)=0$ and $s(0)=0.$ a. Determine and graph the position function, for $t \geq 0.$ b. How far does the racer travel in the first 4 seconds? c. At this rate, how long will it take the racer to travel $\frac{1}{4} \mathrm{mi} ?$ d. How long does it take the racer to travel $300 \mathrm{ft}$ ? e. How far has the racer traveled when it reaches a speed of $178 \mathrm{ft} / \mathrm{s} ?$

Acceleration A drag racer accelerates at $a(t)=88 \mathrm{ft} / \mathrm{s}^{2} .$ Assume that $v(0)=0$ and $s(0)=0.$
a. Determine and graph the position function, for $t \geq 0.$
b. How far does the racer travel in the first 4 seconds?
c. At this rate, how long will it take the racer to travel $\frac{1}{4} \mathrm{mi} ?$
d. How long does it take the racer to travel $300 \mathrm{ft}$ ?
e. How far has the racer traveled when it reaches a speed of $178 \mathrm{ft} / \mathrm{s} ?$

Key Concepts

Solving Equations in Kinematics

Solving equations derived from the integrated motion functions is critical for finding specific information such as the time taken to travel a certain distance or the distance covered at a given time. This process typically involves algebraic manipulation and solving for variables within the context of the kinematic equations.

Initial Conditions and Constants of Integration

When integrating to find velocity or position, initial conditions (such as initial velocity and initial position) are essential to determine the constants of integration. These conditions ensure that the resulting functions accurately describe the motion of an object from a known starting state.

Constant Acceleration Kinematics

This concept involves understanding the motion of objects when the acceleration is constant. In kinematics, constant acceleration allows for the derivation of explicit formulas for velocity and position as functions of time using basic calculus, which are foundational in analyzing uniformly accelerated motion.

Integration to Determine Motion Functions

Integration is used to transition between acceleration, velocity, and position functions. Since acceleration is the derivative of velocity and velocity is the derivative of position, integrating the acceleration function provides the velocity function, and integrating the velocity function yields the position or displacement function, each accompanied by a constant of integration.

Transcript

00:02 Okay, so here we have an acceleration.

00:04 We have a drag racer, and their acceleration is 88 feet per second per second, and we know that at time zero, the velocity and the position are both zero.

00:17 So what we're going to do first is we're going to be figuring out our velocity and our position, the functions for both of those.

00:27 Then we're going to graph our position.

00:30 So if we want to find the velocity, find the anti -derivative of 88, which is 88 t feet per second per second.

00:41 Now because this would have a plus c, but because our velocity starts at zero, we know that our velocity, our c is equal to zero.

00:49 So we have 88t.

00:51 Same thing for this.

00:52 If we do this, we're going to do 44t squared feet per second per second.

00:58 And the same deal happens because of the fact that our time equals zero, our distance is equal to zero, we know that c has to be equal to zero.

01:07 So it would be 44, zero squared, plus c equals zero.

01:12 So c would be equal to zero.

01:21 So from there, we can graph this.

01:23 So 44t squared, if that's the case, at zero we're starting at zero, 44 is going to be one, and it's going to be at 176 at 2.

01:32 And this is going to keep on going.

01:35 A quadratic function.

01:37 So moving on, we're going to move on to the next part.

01:40 It asks, how far does the racer travel in the first four seconds? so to do that, for the next part, we do s of 4 equals 44 t squared.

01:52 So we get 44 times 16, which is equal to 704 feet.

02:11 Okay, and then moving on, it asks us at this rate, how long will it take the racer to travel one -fourth of a mile? so we need to do a bit of conversion here.

02:23 We know, we should know that one -fourth of a mile is equal to, we'll forget that up, a mile is equal to 5 ,280 feet.

02:38 So if we know that, if we divide 5 ,280 divided by 4, we know that 1⁄2m.

02:45 We know that 1 4th of mile equals 13 ,300 and 20 feet.

02:52 So now to figure out how long it will take for a racer to travel that distance, we just make our distance formula, our distance function, equal to 1320...

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