Mathematica Project - Position, Velocity, and Acceleration - plotting

Question

Mathematica Project - Position, Velocity and Acceleration

A rocket is fired vertically upward into the air from a launching platform 100 ft above the ground. The platform is then retracted, and the rocket is allowed to fall to the ground.

The height (in ft) of the rocket relative to the ground t seconds after launch is given by:

s(t)=-16t^2+80t+100

1) Using Mathematica, define a function s(t).
2) Find the average velocity for t=1 ∈[1,2].

3) Find the velocity function, v(t). You'll want to define this as a function in Mathematica as well, so that you can refer to it easily later.

4) Find the instantaneous velocity at t=2s. How does this velocity compare with the average velocity you found in problem 2? Explain.

5) Find the equation of the line tangent to (s)t at t=2, and plot both this line and s(t) on one set of coordinate axes. Is your line actually tangent to the curve? Does the slope look right?

6) When does the rocket reach its highest point, and what is the velocity then?

7) Use Mathematica's NSolve command to find out when the rocket lands. What is its velocity when it hits the ground?

8) Plot s(t) and v(t) together on one set of coordinate axes. What do you notice about v(t) when s(t) is increasing? What about when s(t) is decreasing? At its maximum?

9) Find the acceleration function, a(t). What are the units for acceleration in this problem? Why is a(t) always negative, and what does it tell you about the velocity in this problem?

10) Notice that, on the way up, the rocket's velocity is positive, and its acceleration is negative. Is the rocket speeding up or slowing down? On the way down, both the velocity and acceleration are negative. Is the rocket speeding up or slowing down then?

Answer 1

1) The function is quadratic, describes parabola opened down because its first coefficient is negative (-16).

2) The average speed is defined as a total distance divided by total time.

The total distnace (when 1≤t≤2) is the difference traveled between 1st abd 2nd seconds:

                                   h =  s(2) - s(1) = 32 ft

where s(1) = -16 +80 +100 = 164ft and s(2) = -16x4 +80x2 +100 = 196 ft (direction is still the same).

Thus, the average speed is 32ft/sec

3) The velocity function is a derivative of s(t) and equals

                                              v(t) = -32t + 80

4) Instanteneous speed at t=2 sec is therefore 16 ft/s (-32x2 +80), which is twicw slower compared with

the average speed in 2);

5) Tangent of the line at t=2 sec to the curve s(t) is the value of speed at that moment of time. We just found it

                                                     v(2) = 16 ft/sec

The equation of the tangent line should be like

                                                       y(t) = 16t + b

Because y(2) = s(2) = 196, then b = 196-32 =164 ft. Thus,

                                                     y(t) = 16t +164

6) The rocket reaches at its maximum point after

                       t = -(-80/32) = 2.5 seconds

And maximum height is s _max = 200 ft

7)  It reaches the grond after

                          t = sqrt(2x200/32) = 3.54 seconds

8) The acceleration is always the same = - 32ft/sec² I(directed downward).

9) The rocket is slowing dwon when it moves upward and accelerates when   

goes down (speed and acceleration have the same direction in this case)

You can use you software to plot the graphs required. Good luck!

Answer 2

s(2) = 100 + 80(2) - 16(2)² = 100 + 160 - 64 = 196 ft

s(1) = 100 + 80(1) - 16(1)² = 100 + 80 - 16 = 164 ft

Average Velocity = v_avg = [s(2) - s(1)]/(2 - 1) = 32 ft/s

Power rule: d/dt (tⁿ) = n*t^n-1

v(t) = ds/dt = d/dt (100 + 80t - 16t²) = 80*t⁰ - 2*16*t¹ = 80 - 32t

v(1) = 80 - 32(1) = 48 ft/s

v(2) = 80 - 32(2) = 80 - 64 = 16 ft/s

v_avg = [v(1) + v(2)]/2 =  (48 + 16)/2 = 32 ft/s

Tangent: T(2) = v(2)*(t - 2) + s(2) = 16(t - 2) + 196 = 16t - 32 + 196 = 16t + 164

For maximum altitude, set v(t) equal to 0.

v(t) = 80 - 32t = 0.

t = 80/32 = 2.5 s

Peak Altitude = s_max = s(2.5) = 100 + 80(2.5) - 16(2.5)² = 100 + 200 - 100 = 200 ft

I don't know Mathematics, but I do know that the rocket lands when s(t) = 0.

100 + 80t - 16t² = 0

t_imp = [-80 - sqrt(80² - 4*100*(-16))]/(2*(-16)) = 6.0355 s

Impact Velocity = v(t_imp) = 80 - 32(6.0355) = -113.136 ft/s (downward)

v(t) and speed decreases as s(t) increases. As s(t) decreases, speed increases but v(t) decreases. That is because the rocket is accelerating in the opposite direction.

a(t) = dv/dt = -32 ft/s²

The acceleration is negative because the acceleration is due to gravity. The velocity is decreasing even when the speed increases due to direction. 

The rocket is slowing down while ascending, but speeding up while descending. 

Answer 3

Good hearing from you, Jasha -- here's the breakdown:

height= -16t^2 +80t +100 ... velocity= -32t +80 ,,, acceleration= -32 ... at t=0, height= 100, v= 80, a= -32

Physically at launch, ht is 100ft above ground & initial speed is 80ft/s UP, with constant a= 32ft/s/s DOWN due to gravity ... every second, the velocity negates by 32ft/s ==> 80(0), 48(1), 16(2), zero at peak(2.5s), -32(3.5), -64(4.5), -80(5) when 100ft above ground ... notice after the peak at 2.5s, the minus sign means DOWN as the rocket falls 32ft/s faster each second due to constant gravity.  

Ave velocity= ds/dt where "d" means change ... at t=1, s(1)= 180-16= 164ft, average v(1)= s(1)-s(0)/(t1-t0) Ave velocity= 164-100/1= 64ft/s.

instant v(2)= 80-64= 16 ft/s ... because of gravity, v is slowing ..... the "tangent" of s(t) is the slope  of the height function, which indicates how the ht is changing per second, which is the instant  v(t).

Finally, at t=5, the rocket is still 100ft above ground and falling at 80ft/sec. When does it touch the ground? s(t) should be 0= -16t^2 +80t +100 ... answer would be a little more than 6 seconds ... Regards, sir :)