Math Problem Statement

A particle is moving with a displacement of x(t)=𝑑3+3𝑑2+1 measured in meters. a) Find the average velocity from t=1s to t=3s b) Find an equation for the instantaneous velocity c) Find the instantaneous velocity at t=1s and simplify your answer.

Solution

We are given the displacement function of a particle, x(t)=t3+3t2+1x(t) = t^3 + 3t^2 + 1, where x(t)x(t) is measured in meters and tt is in seconds. Let's solve each part step by step.

a) Find the average velocity from t=1 st = 1 \, \text{s} to t=3 st = 3 \, \text{s}

The formula for the average velocity over an interval [t1,t2][t_1, t_2] is: vavg=x(t2)βˆ’x(t1)t2βˆ’t1v_{\text{avg}} = \frac{x(t_2) - x(t_1)}{t_2 - t_1} In this case, t1=1 st_1 = 1 \, \text{s} and t2=3 st_2 = 3 \, \text{s}. First, let's calculate x(1)x(1) and x(3)x(3).

x(1)=(1)3+3(1)2+1=1+3+1=5 mx(1) = (1)^3 + 3(1)^2 + 1 = 1 + 3 + 1 = 5 \, \text{m} x(3)=(3)3+3(3)2+1=27+27+1=55 mx(3) = (3)^3 + 3(3)^2 + 1 = 27 + 27 + 1 = 55 \, \text{m}

Now, calculate the average velocity: vavg=55βˆ’53βˆ’1=502=25 m/sv_{\text{avg}} = \frac{55 - 5}{3 - 1} = \frac{50}{2} = 25 \, \text{m/s}

Thus, the average velocity from t=1 st = 1 \, \text{s} to t=3 st = 3 \, \text{s} is 25 m/s25 \, \text{m/s}.


b) Find an equation for the instantaneous velocity

The instantaneous velocity is the derivative of the displacement function x(t)x(t) with respect to time tt. Let's find the derivative of x(t)=t3+3t2+1x(t) = t^3 + 3t^2 + 1.

v(t)=dx(t)dt=3t2+6tv(t) = \frac{dx(t)}{dt} = 3t^2 + 6t

Thus, the equation for the instantaneous velocity is: v(t)=3t2+6tv(t) = 3t^2 + 6t


c) Find the instantaneous velocity at t=1 st = 1 \, \text{s} and simplify your answer

To find the instantaneous velocity at t=1 st = 1 \, \text{s}, substitute t=1t = 1 into the equation for v(t)v(t):

v(1)=3(1)2+6(1)=3+6=9 m/sv(1) = 3(1)^2 + 6(1) = 3 + 6 = 9 \, \text{m/s}

Thus, the instantaneous velocity at t=1 st = 1 \, \text{s} is 9 m/s9 \, \text{m/s}.


Summary of Results:

a) The average velocity from t=1 st = 1 \, \text{s} to t=3 st = 3 \, \text{s} is 25 m/s25 \, \text{m/s}.
b) The equation for the instantaneous velocity is v(t)=3t2+6tv(t) = 3t^2 + 6t.
c) The instantaneous velocity at t=1 st = 1 \, \text{s} is 9 m/s9 \, \text{m/s}.

Would you like further details or have any questions about these solutions?

Here are 5 additional related questions:

  1. What would the instantaneous velocity be at t=2 st = 2 \, \text{s}?
  2. How would you find the acceleration of the particle at any time tt?
  3. How does the velocity change between t=1 st = 1 \, \text{s} and t=3 st = 3 \, \text{s}?
  4. Can you find the time when the particle's velocity is zero?
  5. What is the significance of the constant term in the displacement function?

Tip: The derivative of a displacement function gives you the velocity, and the derivative of velocity gives you the acceleration!

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Math Problem Analysis

Mathematical Concepts

Calculus
Kinematics
Differentiation

Formulas

Average velocity: v_avg = (x(t2) - x(t1)) / (t2 - t1)
Instantaneous velocity: v(t) = dx/dt

Theorems

Fundamental Theorem of Calculus

Suitable Grade Level

Grades 11-12