A particle is moving with a displacement of x(t)=𝑡3+3𝑡2+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.


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

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

The formula for the average velocity over an interval \( [t_1, t_2] \) is:
\[
v_{\text{avg}} = \frac{x(t_2) - x(t_1)}{t_2 - t_1}
\]
In this case, \( t_1 = 1 \, \text{s} \) and \( t_2 = 3 \, \text{s} \). First, let's calculate \( x(1) \) and \( x(3) \).

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

Now, calculate the average velocity:
\[
v_{\text{avg}} = \frac{55 - 5}{3 - 1} = \frac{50}{2} = 25 \, \text{m/s}
\]

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

---

### b) Find an equation for the instantaneous velocity

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

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

Thus, the equation for the instantaneous velocity is:
\[
v(t) = 3t^2 + 6t
\]

---

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

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

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

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

---

### Summary of Results:
a) The average velocity from \( t = 1 \, \text{s} \) to \( t = 3 \, \text{s} \) is \( 25 \, \text{m/s} \).  
b) The equation for the instantaneous velocity is \( v(t) = 3t^2 + 6t \).  
c) The instantaneous velocity at \( t = 1 \, \text{s} \) is \( 9 \, \text{m/s} \).

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 \( t = 2 \, \text{s} \)?
2. How would you find the acceleration of the particle at any time \( t \)?
3. How does the velocity change between \( t = 1 \, \text{s} \) and \( t = 3 \, \text{s} \)?
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!

A particle is moving with a displacement of x(t) = t^3 + 3t^2 + 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.

Learn how to calculate the average and instantaneous velocity for a particle moving with the displacement function x(t) = t^3 + 3t^2 + 1. This detailed solution walks through finding average velocity from t = 1s to t = 3s and the equation for instantaneous velocity using differentiation. Suitable for students in Grades 11-12.

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