To find the velocity of a particle, we need to differentiate the displacement equation $s(t)$ with respect to time $t$.

Given: $s = 2t^2$ where $s$ is the displacement in meters and $t$ is the time in seconds.

Step 1: Find the velocity function $v(t)$

The velocity is the derivative of displacement with respect to time: $v(t) = \frac{ds}{dt}$ Differentiate $s = 2t^2$: $v(t) = \frac{d}{dt} (2t^2) = 4t$

Step 2: Calculate the velocity at specific times

Now that we have the velocity function $v(t) = 4t$, we can substitute the given times:

1. At $t = a$: $v(a) = 4a \quad \text{m/s}$

2. At $t = 1$: $v(1) = 4(1) = 4 \quad \text{m/s}$

3. At $t = 2$: $v(2) = 4(2) = 8 \quad \text{m/s}$

4. At $t = 3$: $v(3) = 4(3) = 12 \quad \text{m/s}$

Final Results:

• At $t = a$, $v = 4a$ m/s.
• At $t = 1$, $v = 4$ m/s.
• At $t = 2$, $v = 8$ m/s.
• At $t = 3$, $v = 12$ m/s.

It seems your results for $t = 1, 2, 3$ don't match the correct answers you've shared. Based on the equation $s = 2t^2$, the results calculated above are correct under the assumption that this is the correct equation of motion.

Do you have any questions or would you like further details?

Related Questions:

1. How does differentiation help in calculating velocity from displacement?
2. What is the significance of the second derivative in motion (i.e., acceleration)?
3. How would you find the velocity if the equation was $s = 3t^3$?
4. Can velocity be negative for this particle? Why or why not?
5. How would the motion change if the coefficient of $t^2$ was negative?

Tip:

The velocity function is always the derivative of displacement, and acceleration is always the derivative of velocity!