Straight-Line Motion: Connecting Position, Velocity, and Acceleration

#Straight-Line Motion: Connecting Position, Velocity, and Acceleration 🚗

Hey there, future calculus master! Ready to see how those derivatives you've been acing actually apply to the real world? Let's dive into rectilinear motion, where we'll use derivatives to connect position, velocity, and acceleration. It's like unlocking the secret language of movement! 🚀

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#🌊 Derivatives and Motion

Remember how the derivative gives you the instantaneous rate of change? Well, that's the key! In motion, we use derivatives to see how things are changing right now.

#🛥️ Velocity and Speed

If $x(t)$ is a function that tells you the position of an object at any time $t$, then its derivative, $x'(t)$ (or $v(t)$), gives you the velocity at that exact moment. Velocity is all about the rate of change of position with respect to time. It's also signed, meaning it tells you the direction of motion:

• When $v(t)$ is negative, the object is moving left: ➖ = ⬅️.
• When $v(t)$ is positive, the object is moving right: ➕ = ➡️.

#🚤 Acceleration

Similarly, the derivative of the velocity function, $v'(t)$ (or $a(t)$), gives you the acceleration of the object. Acceleration is the rate at which velocity changes. And guess what? Since $v(t) = x'(t)$, that means $a(t) = x''(t)$! The second derivative of position is acceleration! 🤯

• If acceleration is positive, the velocity is increasing in the direction of motion (speeding up).
• If acceleration is negative, the velocity is decreasing in the direction of motion (slowing down).
• If acceleration is zero, the object is moving at a constant velocity.

Image Courtesy of BYJU’S

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#📝 Rectilinear Motion: Practice

Time to put your knowledge to the test! Let's work through some problems.

#❓ Rectilinear Motion: Problems

#Question 1:

A particle moves along the x-axis. The function $x(t)$ gives the particle’s position at any time $t >= 0$. $$x(t) = 13t - 9$$ What is the particle’s acceleration $a(t)$ at $t=3$?

#Question 2:

A particle moves along the x-axis. The function $x(t)$ gives the particle’s position at any time $t >= 0$. $$x(t) = 4t^2 - 3t + 16$$ What is the particle’s velocity $v(t)$ at $t=4$?

#✅ Rectilinear Motion: Solutions

#Question 1:

Acceleration is the second derivative of position. So, let's find $x''(t)$:

$a(t) = x''(t) = \frac{d}{dt}[\frac{d}{dt}[13t-9]] = 0$

Since the acceleration is constant at 0 for all $t >= 0$, the particle’s acceleration at $t=3$ is also 0. #### Question 2:

Velocity is the first derivative of position. So, let's find $x'(t)$:

$v(t) = x'(t) = \frac{d}{dt}[4t^2 - 3t + 16] = 8t - 3$

The particle's velocity is $v(t) = 8t - 3$. Plugging in $t=4$, we get $v(4) = 8(4) - 3 = 29$.

Practice Question

{ "multiple_choice": [ { "question": "A particle moves along the x-axis with position given by $x(t) = t^3 - 6t^2 + 9t + 1$. For what values of $t$ is the particle moving to the left?", "options": [ "$0 < t < 1$", "$1 < t < 3$", "$t > 3$", "$t < 0$ or $t > 3$", "$t < 1$" ], "answer": "$1 < t < 3$" }, { "question": "The velocity of a particle moving along the x-axis is given by $v(t) = 2t - 6$. At what time $t$ is the particle at rest?", "options": [ "0", "1", "2", "3", "4" ], "answer": "3" }, { "question": "A particle moves along the x-axis such that its position is given by $x(t) = t^2 - 4t + 3$. What is the average velocity of the particle from $t = 1$ to $t = 3$?", "options": [ "-2", "-1", "0", "1", "2" ], "answer": "0" } ], "free_response": { "question": "A particle moves along the x-axis such that its velocity is given by $v(t) = t^2 - 4t + 3$. The position of the particle at time $t=0$ is $x(0) = 5$.", "parts": [ { "part": "(a)", "instruction": "Find the acceleration of the particle at time $t=2$.", "solution": "$a(t) = v'(t) = 2t - 4$. $a(2) = 2(2) - 4 = 0$.", "points": "1 point for finding the derivative, 1 point for the correct answer" }, { "part": "(b)", "instruction": "Find all times $t$ in the interval $[0, 5]$ when the particle is at rest.", "solution": "The particle is at rest when $v(t) = 0$. $t^2 - 4t + 3 = (t-1)(t-3) = 0$. Thus, $t = 1$ and $t = 3$.", "points": "1 point for setting $v(t) = 0$, 1 point for finding both correct times" }, { "part": "(c)", "instruction": "Find the position of the particle at time $t=3$.", "solution": "$x(t) = \\int v(t) dt = \\int (t^2 - 4t + 3) dt = \\frac{1}{3}t^3 - 2t^2 + 3t + C$. Since $x(0) = 5$, we have $C = 5$. So, $x(t) = \\frac{1}{3}t^3 - 2t^2 + 3t + 5$. Then, $x(3) = \\frac{1}{3}(3)^3 - 2(3)^2 + 3(3) + 5 = 9 - 18 + 9 + 5 = 5$.", "points": "2 points for finding the antiderivative, 1 point for using initial condition, 1 point for the correct position" } ] } }