Acceleration Equations: Solving For Velocity & Distance

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Hey everyone! Let's dive into a fun physics problem involving acceleration, velocity, and distance. We're given an equation for acceleration and need to figure out a few things. This is a classic example of how calculus, specifically integration, helps us understand motion. So, let's break it down step-by-step.

Understanding the Problem

We're dealing with a scenario where the acceleration of an object isn't constant; it changes with time. We're given the equation for acceleration as a function of time:

a = 4t^4 + 7t^3 + 2

Where:

  • a represents acceleration (in meters per second squared, m/s²)
  • t represents time (in seconds, s)

Our mission, should we choose to accept it (and we do!), is to find:

  • The acceleration at a specific time (t = 3 s)
  • The velocity at another specific time (t = 2.5 s)
  • The total distance traveled within a time interval (from t = 0 s to t = 3 s)

To do this, we'll be using the fundamental relationship between acceleration, velocity, and displacement, which involves integration. Remember, velocity is the integral of acceleration with respect to time, and displacement (change in position, related to distance) is the integral of velocity with respect to time. These are key concepts, so let's keep them in mind as we proceed.

a) Acceleration at t = 3 s

Okay, this is the easiest part! To find the acceleration at t = 3 seconds, we simply substitute t = 3 into our acceleration equation:

a = 4(3)^4 + 7(3)^3 + 2

Let's calculate this:

a = 4 * 81 + 7 * 27 + 2
a = 324 + 189 + 2
a = 515 m/s²

So, the acceleration at t = 3 seconds is a whopping 515 m/s². That's some serious acceleration! This part highlights a core concept: instantaneous acceleration. We're finding the acceleration at a specific instant in time, not an average over an interval.

b) Velocity at t = 2.5 s

Now, things get a bit more interesting. To find the velocity, we need to integrate the acceleration function with respect to time:

v = ∫ a dt = ∫ (4t^4 + 7t^3 + 2) dt

Remember the power rule for integration? ∫tⁿ dt = (tⁿ⁺¹)/(n+1) + C, where C is the constant of integration. Let's apply that:

v = (4t⁵)/5 + (7t⁴)/4 + 2t + C

Here's where that C comes in. It's the constant of integration, and it's crucial! It represents the initial velocity of the object. Without more information (like the velocity at t=0), we can't determine the exact value of C. For now, let's assume the object starts from rest, meaning its initial velocity is 0. So, C = 0.

Now we have:

v = (4t⁵)/5 + (7t⁴)/4 + 2t

To find the velocity at t = 2.5 seconds, we plug in t = 2.5:

v = (4(2.5)⁵)/5 + (7(2.5)⁴)/4 + 2(2.5)

Calculating this gives us:

v ≈ (4 * 97.66)/5 + (7 * 39.06)/4 + 5
v ≈ 78.125 + 68.359 + 5
v ≈ 151.484 m/s

So, the velocity at t = 2.5 seconds is approximately 151.484 m/s. This showcases how integration allows us to move from acceleration to velocity. We're essentially summing up the effect of acceleration over time to get the velocity. It's like adding up all the little pushes and shoves to see how fast the object is ultimately going.

c) Distance Traveled from t = 0 s to t = 3 s

This is the final boss! To find the distance traveled, we need to integrate the velocity function with respect to time:

x = ∫ v dt = ∫ [(4t⁵)/5 + (7t⁴)/4 + 2t] dt

Again, using the power rule for integration:

x = (4t⁶)/(5*6) + (7t⁵)/(4*5) + t² + D
x = (2t⁶)/15 + (7t⁵)/20 + t² + D

We have another constant of integration, D. This represents the initial position of the object. If we assume the object starts at the origin (x = 0) at t = 0, then D = 0. So,

x = (2t⁶)/15 + (7t⁵)/20 + t²

To find the distance traveled from t = 0 s to t = 3 s, we need to calculate the position at t = 3 s and subtract the position at t = 0 s. Since we've assumed the initial position is 0, we just need to calculate x at t = 3:

x = (2(3)⁶)/15 + (7(3)⁵)/20 + (3)²
x = (2 * 729)/15 + (7 * 243)/20 + 9
x = 97.2 + 85.05 + 9
x ≈ 191.25 m

Therefore, the distance traveled by the object from t = 0 s to t = 3 s is approximately 191.25 meters. This part brings it all together. We started with acceleration, integrated to find velocity, and then integrated velocity to find position (and thus, distance). This illustrates the hierarchical relationship between these kinematic quantities and how calculus allows us to move between them.

d) The Integral ∫ 5t³ dt

Let's tackle this integral separately, as it was presented as an example. It's a straightforward application of the power rule:

∫ 5t³ dt = 5 ∫ t³ dt

Applying the power rule:

= 5 * (t⁴/4) + C
= (5t⁴)/4 + C

So, the integral of 5t³ dt is (5t⁴)/4 + C, where C is the constant of integration. This is a good reminder of the basic integration techniques that underpin the solutions to the more complex problems we tackled earlier.

Key Takeaways

Alright, guys, that was quite a journey! Let's recap what we've learned:

  • Acceleration, velocity, and displacement are related through calculus. Velocity is the integral of acceleration, and displacement is the integral of velocity.
  • Integration requires a constant of integration (C). This constant represents the initial condition (e.g., initial velocity or initial position) and is crucial for finding a specific solution.
  • The power rule of integration (∫tⁿ dt = (tⁿ⁺¹)/(n+1) + C) is your friend! Master this, and you'll be able to solve a wide range of physics problems.
  • Understanding the physical meaning of integration is key. We're essentially summing up the effect of a quantity over time to get another quantity.

By working through this problem, we've not only found the answers but also reinforced some fundamental physics and calculus concepts. Keep practicing, and you'll become a master of motion!