Math Problem Statement

The velocity of a particle moving in a straight line is given by v(t) = 5t + 18t^2. If s(3) = 8, where s(t) is the distance traveled, find s(5)

Solution

To solve this problem, we need to find s(5)s(5) for the given velocity function v(t)=5t+18t2v(t) = 5t + 18t^2, and the initial condition s(3)=8s(3) = 8.

Steps:

  1. Relation Between Velocity and Distance: Velocity v(t)v(t) is the derivative of the distance function s(t)s(t): v(t)=ds(t)dt.v(t) = \frac{ds(t)}{dt}.

    Therefore, the distance function s(t)s(t) is obtained by integrating v(t)v(t): s(t)=v(t)dt.s(t) = \int v(t) \, dt.

  2. Integrate the Velocity Function: Given v(t)=5t+18t2v(t) = 5t + 18t^2, integrate it: s(t)=(5t+18t2)dt=5t22+18t33+C,s(t) = \int (5t + 18t^2) \, dt = \frac{5t^2}{2} + \frac{18t^3}{3} + C, where CC is the constant of integration. Simplify: s(t)=5t22+6t3+C.s(t) = \frac{5t^2}{2} + 6t^3 + C.

  3. Apply the Initial Condition s(3)=8s(3) = 8: Substitute t=3t = 3 and s(3)=8s(3) = 8 into the equation to solve for CC: 8=5(3)22+6(3)3+C.8 = \frac{5(3)^2}{2} + 6(3)^3 + C. Simplify the terms: 8=592+627+C,8 = \frac{5 \cdot 9}{2} + 6 \cdot 27 + C, 8=452+162+C.8 = \frac{45}{2} + 162 + C. Combine: 8=452+3242+C,8 = \frac{45}{2} + \frac{324}{2} + C, 8=3692+C.8 = \frac{369}{2} + C. Solve for CC: C=83692=1623692=3532.C = 8 - \frac{369}{2} = \frac{16}{2} - \frac{369}{2} = -\frac{353}{2}.

    Thus, the distance function is: s(t)=5t22+6t33532.s(t) = \frac{5t^2}{2} + 6t^3 - \frac{353}{2}.

  4. Find s(5)s(5): Substitute t=5t = 5 into s(t)s(t): s(5)=5(5)22+6(5)33532.s(5) = \frac{5(5)^2}{2} + 6(5)^3 - \frac{353}{2}. Simplify each term: s(5)=5252+61253532,s(5) = \frac{5 \cdot 25}{2} + 6 \cdot 125 - \frac{353}{2}, s(5)=1252+7503532.s(5) = \frac{125}{2} + 750 - \frac{353}{2}. Combine terms: s(5)=1253532+750,s(5) = \frac{125 - 353}{2} + 750, s(5)=2282+750,s(5) = -\frac{228}{2} + 750, s(5)=114+750.s(5) = -114 + 750. s(5)=636.s(5) = 636.

Final Answer:

s(5)=636.s(5) = 636.

Would you like more details or have any questions?


Related Questions:

  1. How would the solution change if s(3)s(3) had a different value?
  2. Can you explain how the integral is related to velocity and displacement?
  3. What would s(t)s'(t) represent in this context?
  4. How would the function change if v(t)v(t) were instead 5t+18t2+35t + 18t^2 + 3?
  5. What if the particle moved in a circular path instead of a straight line?

Tip:

Always check your integration constants carefully when solving problems involving definite or indefinite integrals.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Velocity
Distance
Integral Calculus
Differentiation

Formulas

v(t) = ds(t)/dt
s(t) = ∫v(t) dt

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12