Question:
A particle moving along the x-axis has a position given by x=(24t - 2.0t³)m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero
Answer:
24 m/s
Explanation:
Given:
x=(24t - 2.0t³)m
First find velocity function v(t):
v(t) = ẋ(t) = 24 - 2*3t²
v(t) = ẋ(t) = 24 - 6t²
Find the acceleration function a(t):
a(t) = Ẍ(t) = V(t) = -6*2t
a(t) = Ẍ(t) = V(t) = -12t
At acceleration = 0, take time as T in velocity function.
0 =v(T) = 24 - 6T²
Solve for T
Substitute -2 for t in acceleration function:
a(t) = a(T) = a(-2) = -12(-2) = 24 m/s
Acceleration = 24m/s
Answer:
There is an interval of 24.28s in which the rocket is above the ground.
Explanation:





From Kinematics, the position
as a function of time when the engine still works will be:

At what time the altitud will be
?
⇒ 
Using the quadratic formula:
.
How much time does it take for the rocket to touch the ground? No the function of position is:

Where our new initial position is
, the velocity when the engine breaks is
and the only acceleration comes from gravity (which points down).
Now, when the rocket tounches the ground:
Again, using the quadratic ecuation:

Now, the total time from the moment it takes off and the moment it tounches the ground will be:
.