Question

A model rocket is launched straight upward. The rocket's speed is 98.0 3.0 m/s when its engine burns out. The engine only burns for 2.6 s.

a) What is the acceleration of the rocket during this time?
b) How high does the rocket fly while the engine is burning ?

Answer 1

The acceleration of the rocket during the burning time of the engine is approximately 36.538 m/s². The rocket reaches a height of approximately 82.545 meters while the engine is burning.

Step-by-step explanation:

To solve this problem, we can use the equations of motion for constant acceleration. Let's start with part (a) to find the acceleration of the rocket during the burning time of the engine. We can use the equation:

v = u + at

where v is the final velocity (98.0 m/s), u is the initial velocity (3.0 m/s), a is the acceleration (unknown), and t is the time (2.6 s).

Substituting the known values into the equation:

98.0 = 3.0 + a × 2.6

Simplifying the equation, we have:

a = (98.0 - 3.0)/2.6

Calculating the acceleration, we get

a ≈ 36.538 m/s²

For part (b), we can use the equation for displacement:

s = ut + 0.5at²

where s is the displacement (unknown), u is the initial velocity (3.0 m/s), a is the acceleration (36.538 m/s²), and t is the time (2.6 s).

Substituting the given values into the equation:

s = 3.0 × 2.6 + 0.5 × 36.538 × (2.6)²

Simplifying the equation, we have:

s ≈ 82.545 meters