To find the velocity of the particle at time $t = a$, we first need to differentiate the position function $s(t) = 2t^2 + 4t - 5$ with respect to time $t$.

The velocity function $v(t)$ is the derivative of the position function $s(t)$:

$v(t) = \frac{d}{dt}[2t^2 + 4t - 5]$

Differentiating each term:

• The derivative of $2t^2$ is $4t$.
• The derivative of $4t$ is $4$.
• The derivative of $-5$ is $0$.

Thus, the velocity function is:

$v(t) = 4t + 4$

Now, to find the velocity at $t = 1$ second, $t = 2$ seconds, and $t = 3$ seconds, we substitute these values into the velocity function:

• At $t = 1$:

$v(1) = 4(1) + 4 = 4 + 4 = 8 \, \text{m/s}$

• At $t = 2$:

$v(2) = 4(2) + 4 = 8 + 4 = 12 \, \text{m/s}$

• At $t = 3$:

$v(3) = 4(3) + 4 = 12 + 4 = 16 \, \text{m/s}$

Summary of velocities:

• $v(1) = 8 \, \text{m/s}$
• $v(2) = 12 \, \text{m/s}$
• $v(3) = 16 \, \text{m/s}$

Would you like to explore further or need clarification on any step?

Here are some related questions:

1. How would you calculate the acceleration of the particle at time $t = a$?
2. What is the position of the particle at time $t = 1$, $t = 2$, and $t = 3$?
3. What is the particle's displacement between $t = 1$ and $t = 3$?
4. How would the velocity change if the position function were $s = 3t^2 + 2t - 7$?
5. What is the total distance traveled by the particle from $t = 0$ to $t = 3$?

Tip: When finding the velocity or acceleration, always check if the position function is quadratic (or higher) since the derivatives of polynomials are easier to calculate!