Model rocket (velocity and time equation help please)

[Duke77]

A model rocket is fired vertically upward from rest. Its acceleration for the first three seconds is a(t)=60t, at which time the fuel is exhausted and it becomes a freely “falling” body. Fourteen seconds after the fuel is exhausted, the rocket’s parachute opens, and the (downward) velocity slows linearly to -18 ft/sec in 5 seconds. The rocket then “floats” to the ground at that rate. Find the position function s and the velocity function v for any time t, then sketch the graphs of s and v.

This is the whole question and I don't expect help with the graphs but if I could get some help on the problem I would appreciate it greatly.

 

[Doc Al]

one step at a time

Attack it step by step.

For T = 0, 3 secs: Integrate the acceleration function (twice) to find expressions for v and s.
For T = 3, 14 secs: It's a projectile. Uniform acceleration due to gravity.
For T = 14, 19 secs: the acceleration is given
For T > 19 secs: uniform speed

It's a pain in the butt, but you can do it!

 

[M98Ranger]

To find the position function s and velocity function v for any time t, we can use the equations of motion:

s(t) = s0 + v0t + (1/2)at^2

v(t) = v0 + at

Where s0 is the initial position, v0 is the initial velocity, a is the acceleration, and t is the time.

In this case, we are given that the rocket is fired from rest, so s0 = 0 and v0 = 0. We can also determine the acceleration for the first three seconds using the given information:

a(t) = 60t

a(3) = 60(3) = 180 ft/sec^2

So for the first three seconds, the position function is:

s(t) = (1/2)(180)t^2 = 90t^2 ft

And the velocity function is:

v(t) = (180)t ft/sec

After the fuel is exhausted, the rocket becomes a freely falling body, so the acceleration due to gravity is -32 ft/sec^2. We are given that the parachute opens after 14 seconds, so we can use this information to find the velocity at that time:

v(14) = (180)(14) - (32)(14) = 1960 - 448 = 1512 ft/sec

For the next 5 seconds, the velocity slows linearly to -18 ft/sec, so we can use the slope formula to find the acceleration during this time:

a = (vf - vi)/(tf - ti) = (-18 - 1512)/(5 - 14) = 1494/-9 = -166 ft/sec^2

Using this acceleration, we can find the position function for the next 5 seconds:

s(t) = 1512(5) + (1/2)(-166)(5)^2 = 7550 ft

Finally, after the rocket reaches a velocity of -18 ft/sec, it continues to fall at a constant rate until it reaches the ground. This means that the velocity function is constant at -18 ft/sec for the remaining time. We can use this information to find the position function for the last part of the flight:

s(t) = 7550 + (-18)(t-19) = 7550 - 18t + 342 = -18t + 7892 ft

Therefore,