Math Problem Statement

Solve

1. Find the value of t for which the velocity is increasing if s = 5 - 12t^2 + t^3. 2. Find the velocity when the acceleration is 2 if s = t^3 - t^2.

Solution

Let's solve each problem one by one from the uploaded image:

Problem 1

Given: s=512t2+t3s = 5 - 12t^2 + t^3 We need to find the value of tt for which the velocity is increasing.

  1. Find the Velocity vv: The velocity vv is the first derivative of ss with respect to tt: v=dsdt=ddt(512t2+t3)=24t+3t2v = \frac{ds}{dt} = \frac{d}{dt}(5 - 12t^2 + t^3) = -24t + 3t^2

  2. Find the Acceleration aa: Acceleration aa is the derivative of vv with respect to tt: a=dvdt=ddt(24t+3t2)=24+6ta = \frac{dv}{dt} = \frac{d}{dt}(-24t + 3t^2) = -24 + 6t

  3. Condition for Velocity to Increase: For the velocity to be increasing, the acceleration must be positive: a>024+6t>0a > 0 \Rightarrow -24 + 6t > 0 Solving for tt: 6t>246t > 24 t>4t > 4 Therefore, the velocity is increasing for t>4t > 4.

Problem 2

Given: s=t3t2s = t^3 - t^2 We need to find the velocity when the acceleration is 22.

  1. Find the Velocity vv: Velocity vv is the first derivative of ss with respect to tt: v=dsdt=ddt(t3t2)=3t22tv = \frac{ds}{dt} = \frac{d}{dt}(t^3 - t^2) = 3t^2 - 2t

  2. Find the Acceleration aa: Acceleration aa is the derivative of vv with respect to tt: a=dvdt=ddt(3t22t)=6t2a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 2t) = 6t - 2

  3. Set Acceleration Equal to 2: We are given that the acceleration is 22, so: 6t2=26t - 2 = 2 Solving for tt: 6t=46t = 4 t=23t = \frac{2}{3}

  4. Find the Velocity at t=23t = \frac{2}{3}: Substitute t=23t = \frac{2}{3} into the expression for vv: v=3(23)22(23)v = 3\left(\frac{2}{3}\right)^2 - 2\left(\frac{2}{3}\right) v=34943v = 3 \cdot \frac{4}{9} - \frac{4}{3} v=12943=12129=0v = \frac{12}{9} - \frac{4}{3} = \frac{12 - 12}{9} = 0

    Thus, the velocity is 00 when the acceleration is 22.


Do you need further details on any of these steps?

Additional Questions:

  1. What is the physical interpretation of acceleration being positive or negative?
  2. How does the sign of velocity relate to the direction of motion?
  3. Could we find the time when the particle stops momentarily using these equations?
  4. What would happen to the position function if we modified it slightly, like s=512t2+2t3s = 5 - 12t^2 + 2t^3?
  5. How can we interpret the derivative of acceleration, and what does it represent physically?

Tip:

When analyzing motion, remember that a positive acceleration does not always mean an increase in speed; it depends on the direction of velocity as well.

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Math Problem Analysis

Mathematical Concepts

Calculus
Derivatives
Rectilinear Motion
Velocity
Acceleration

Formulas

Velocity v = ds/dt
Acceleration a = dv/dt

Theorems

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Suitable Grade Level

Grades 11-12