Final answer:
The rocket has an acceleration of 17.78 m/s² upward, and at 3.0 seconds, its speed is approximately 53.34 m/s.
Step-by-step explanation:
To find the magnitude and direction of the rocket's acceleration, we can use the kinematic equation for motion with constant acceleration:
\[ s = ut + \frac{1}{2}at^2 \]
In this formula, s represents the displacement (height in this case), u is the initial velocity (which we assume to be 0 since it starts from the launch pad), t is the time, and a is the acceleration. Knowing that the rocket reached a height of 80.0 m after 3.0 s, we can substitute these values into the equation:
\[ 80.0 = 0 \cdot 3.0 + \frac{1}{2}a(3.0)^2 \]
Solving for a gives us:
\[ a = \frac{2 \cdot 80.0}{3.0^2} \Rightarrow a \approx 17.78 \text{ m/s}^2 \]
The direction of the acceleration is upward, the same as the direction of motion. To find its speed at this time, use theequation:
\[ v = u + at \]
Here, v is the final velocity after time t. So the speed is:
\[ v = 0 + 17.78 \cdot 3.0 \Rightarrow v \approx 53.34 \text{ m/s} \]
The rocket has an acceleration of 17.78 m/s2, and at 3.0 s, its speed is approximately 53.34 m/s.