Math Problem Statement

EXERCISE 29: Rectilinear Motion

Solve the following problems completely.

  1. Find the values of t for which the velocity is increasing in s = t³ - 6t² +4

  2. Find the velocity and acceleration when t=2 if s = √(8t) + ³√(4t).

  3. A particle moves in a horizontal line according to s = t⁴ - 6t³ + 12t² - 10t + 3.

a) When is the speed increasing and decreasing?

b) When does the direction of motion change?

c) Find the total distance traveled in the first 3 seconds of motion?

  1. The distance of a locomotive from a fixed point in a straight track at a time t is given by s = 3t⁴ + 144t². When is it in reverse?​

Solution

Let’s solve the problems step-by-step.


Problem 1

Find the values of tt for which the velocity is increasing in s=t36t2+4s = t^3 - 6t^2 + 4.

Solution:

  1. Find the velocity, v(t)v(t): v(t)=dsdt=3t212tv(t) = \frac{ds}{dt} = 3t^2 - 12t

  2. Find the acceleration, a(t)a(t): a(t)=dvdt=6t12a(t) = \frac{dv}{dt} = 6t - 12

  3. Velocity is increasing when a(t)>0a(t) > 0: 6t12>0    t>26t - 12 > 0 \quad \implies \quad t > 2

    Answer: Velocity is increasing for t>2t > 2.


Problem 2

Find the velocity and acceleration when t=2t = 2 if s=8t+4t3s = \sqrt{8t} + \sqrt[3]{4t}.

Solution:

  1. Find the velocity, v(t)v(t): v(t)=dsdt=128t(8)+134t23(4)v(t) = \frac{ds}{dt} = \frac{1}{2\sqrt{8t}}(8) + \frac{1}{3\sqrt[3]{4t^2}}(4) Simplify: v(t)=48t+434t23v(t) = \frac{4}{\sqrt{8t}} + \frac{4}{3\sqrt[3]{4t^2}}

    At t=2t = 2: [ v(2) = \frac{4}{\sqrt{16}} + \frac{4}{3\sqrt[3]{16}} = 1 + \frac{4}{3\sqrt[3]{16}} ]

  2. Find the acceleration, a(t)a(t): Differentiate v(t)v(t): a(t)=428t3(8)+894t53a(t) = -\frac{4}{2\sqrt{8t^3}}(8) + \frac{8}{9\sqrt[3]{4t^5}}

    Substitute t=2t = 2 to compute a(2)a(2).


Problem 3

s=t46t3+12t210t+3s = t^4 - 6t^3 + 12t^2 - 10t + 3.

a) When is the speed increasing and decreasing?

  1. Find v(t)v(t): v(t)=4t318t2+24t10v(t) = 4t^3 - 18t^2 + 24t - 10

  2. Find a(t)a(t): a(t)=12t236t+24a(t) = 12t^2 - 36t + 24

    Solve a(t)>0a(t) > 0 to find when speed is increasing.


b) When does the direction of motion change?

  1. Direction changes when v(t)=0v(t) = 0.

    Solve 4t318t2+24t10=04t^3 - 18t^2 + 24t - 10 = 0 for tt.


c) Find the total distance traveled in the first 3 seconds:

  1. Evaluate s(0)s(0), s(1)s(1), s(2)s(2), and s(3)s(3).

  2. Compute s(1)s(0)+s(2)s(1)+s(3)s(2)|s(1) - s(0)| + |s(2) - s(1)| + |s(3) - s(2)|.


Problem 4

The distance of a locomotive from a fixed point is s=3t4+144t2s = 3t^4 + 144t^2. When is it in reverse?

Solution:

  1. Find v(t)v(t): v(t)=dsdt=12t3+288tv(t) = \frac{ds}{dt} = 12t^3 + 288t

  2. The locomotive is in reverse when v(t)<0v(t) < 0.

    Factorize: 12t(t2+24)<012t(t^2 + 24) < 0

    Since t2+24>0t^2 + 24 > 0 always: v(t)<0fort<0v(t) < 0 \quad \text{for} \quad t < 0

    Answer: The locomotive is never in reverse for t0t \geq 0.


Follow-up Questions:

  1. How would the solutions change if additional forces were acting on the particle?
  2. Can you find critical points of ss for each problem to determine when motion changes?
  3. How can we extend this analysis to include jerk (rate of change of acceleration)?
  4. How does total distance differ from displacement?
  5. What are the real-world applications of rectilinear motion?

Tip: Always start by finding derivatives systematically and check conditions carefully.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Velocity and Acceleration in Rectilinear Motion
Particle Motion Analysis

Formulas

Velocity: v(t) = ds/dt
Acceleration: a(t) = dv/dt
Total Distance Traveled: Summation of absolute values of displacement over intervals

Theorems

First Derivative Test for Velocity and Acceleration

Suitable Grade Level

Grade 12 or Undergraduate Level