Cars A and B travel in a straight line. The distance of A...

Cars $A$ and $B$ travel in a straight line. The distance of $A$ from the starting point is given as a function of time by $x_A(t) = \alpha{t} + \beta{t}^2$, with $\alpha =$ 2.60 m/s and $\beta =$ 1.20 m/s$^2$. The distance of $B$ from the starting point is $x_B(t) = \gamma{t}^2 - \delta{t}^3$, with $\gamma =$ 2.80 m/s$^2$ and $\delta =$ 0.20 m/s$^3$. (a) Which car is ahead just after the two cars leave the starting point? (b) At what time(s) are the cars at the same point? (c) At what time(s) is the distance from $A$ to $B$ neither increasing nor decreasing? (d) At what time(s) do $A$ and $B$ have the same acceleration?

Key Concepts

Solving Polynomial Equations

Many kinematics problems, particularly when positions or velocities are expressed as polynomial functions of time, require solving polynomial equations. These equations are solved to find the specific times when particular events occur, such as the objects meeting or having matching accelerations.

Relative Motion

Relative motion examines the movement of one object with respect to another by comparing their respective positions, velocities, or accelerations. This concept helps to determine situations where the distance between objects does not change, indicating identical instantaneous velocities, or when two objects are momentarily at the same location.

Kinematics in One Dimension

This concept involves studying the motion of objects moving along a straight path where parameters such as displacement, velocity, and acceleration are functions of time. It provides the framework to describe how an object’s position changes with time and how its speed and direction evolve over time.

Differentiation in Motion Analysis

Differentiation is used to obtain velocity from the position-time function and acceleration from the velocity-time function. By taking derivatives, one can determine instantaneous rates of change, which are crucial in analyzing when two moving objects are at the same position, when relative distances are changing, or when they share the same acceleration.

Transcript

00:01 So part of this problem is asking which position is larger for really small values of t.

00:08 And a way you can do this is by plugging in really small value of t, such as 0 .01, and then just crunching the numbers in your calculator and seeing which one of these values is larger.

00:19 Or you can notice that we have larger powers of t for b than we do for a.

00:26 And so if we plugged in something like 1 -100th of a second, these numbers are going to be really small because one one hundredth squared is a really small number and one one one hundredth cube is an extremely small number also we have a minus sign here so we're going to be subtracting two small numbers whereas for this one we're adding two bigger numbers and so shortly after zero t is equal to zero xb is going to be smaller but then you'll see if t is equal to 100 seconds, then these higher powers start to make b much larger than a.

01:06 And so i just like to solve this by intuition.

01:08 And so by the reasoning i just said, b is a shorter distance away just after t is equal to 0.

01:34 Our b is asking at what time are the two distances equal? and so to solve this, we just equate our two formulas for the positions, xa and xb.

01:49 And so for xa, i have this, and i'm equating it to xb, which is this.

02:02 For these problems, where they give you a bunch of constants, i find it easiest to just leave them in terms of these parameters, alpha, beta, gamma, delta, rather than plug them in.

02:10 It just makes it go a little quicker.

02:13 At this point, we can cancel t.

02:15 And then we have a form for a quadratic equation so that we can use the quadratic formula.

02:22 The standard form of that equation is this.

02:26 Beta minus gamma times t plus alpha is equal to zero.

02:34 And so quadratic formula tells us that t is equal to gamma minus beta plus or minus a square root of beta minus gamma squared minus four delta alpha and this is all over two times a which in this case is two delta now at this point you do actually have to plug in what these constants represent i've written them down here these four numbers here whenever you do that and crunch the number numbers through your calculator you get that t is equal to 2 .275 seconds and t is equal to 5 .725 seconds and so these are the two times where the two x values for a and b are equivalent and that's part b part c asks us at which point is the distance between them not changing and so the way i did this is i wrote down the equation for the distance between them which is equal to xa minus xb but xa and xb already have forms and so plugging in those forms i get alpha t plus beta t squared minus gamma t squared plus delta t cubed this is a function of t and it gives us the distance if you plug in a value of t and then crunch the numbers you'll get the distance between a and b and we want to know when this is not changing and so we have to take its derivative and set that equal to zero so the derivative i'm just going to denote by distance prime is equal to alpha plus 2 beta t minus 2 gamma t plus 3 delta t squared.

04:42 And we need to set this equal to zero to find out when it's not changing.

04:46 And so when we set that equal to zero, we once again get a quadratic because we have t squared.

04:52 And then we have a large linear term here and then our constant term.

04:57 And so we can just plug this into the quadratic formula again.

04:59 I'm not going to write out the whole step at this point...

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