1) The acceleration of a particle is given by ax(t)=? 2.10m/s2 +( 3.03m/s3 )t. a
ID: 2273432 • Letter: 1
Question
1) The acceleration of a particle is given by ax(t)=? 2.10m/s2 +( 3.03m/s3 )t.
a) Find the initial velocity v0x such that the particle will have the same x-coordinate at time t= 3.94s as it had at t=0t=10,
......... m/s
b) What will be the velocity at time t = 3.94s ? .......... m/s
2) A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 6.0m/s2 . Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 11.0s , Powers shuts off the engine and steps out of the helicopter. Assume that the helicopter is in free fall after its engine is shut off, and ignore the effects of air resistance.
a)
Explanation / Answer
1.
Remember that acceleration is the second derivative of the position function and that velocity is the first derivative. So integrating:
Vx(t) = -2.10 t + 1.515t^2 + Vo
X(t) = -1.05 t^2 + 0.505 t^3 + Vo t + Xo
At t =0
X(t) = Xo
At t = 3.94:
X(3.94) = -1.05(3.94)^2 + 0.505(3.94)^3 + Vo(3.94) = Xo
14.58 + 3.94 Vo = Xo
So Vo = (Xo - 14.58)/3.94
Assuming Xo to be zero
Vo = -3.7 m/s <===========Answer to part A
Vx(t) = -2.1 t + 1.515t^2 + Vo
V(3.94) = -2.1(3.94) + 1.515(3.94)^2 - 3.7
V(3.94) = +11.226 m/s
2.
Pretty much a matter of just taking it step by step...
For the initial step, assume it starts at rest at the ground and find the helicopter's height at the top of its trajectory:
d = ut + 1/2 at^2 (u = 0 so the first term vanishes) = 1/2*6*11^2 = 363 m.
The helicopter then free-falls from that height, where its velocity can be assumed to be either 0 or some velocity upward. The question is slightly ambiguous, but lets assume it's the latter, which is the more difficult case. If it's 0 you should be able to follow the same steps but sub in that value and it will be simpler.
First to find the upward velocity of the helicopter when the engine is turned off:
v = u + at (u=0) = 6*11 = 66 m/s upward.
To find the time taken for the helicopter to read the ground, use u = 66 m/s and a = -9.8 m/s/s (since it's in the opposite direction, for a starting distance we already know is 363 m, but treat it as -363 m since it is in the opposite direction to the initial velocity.
d = ut + 1/2 a t^2
-363 = 66t - 4.9 t^2
This is a quadratic equation in t, and will have two roots: you'll have to use common sense and your physics intuition to figure out that it's the one with a positive time that you want!
Too messy to solve a quadratic here, but it ends up being 17.66 s. That's the time from when the engine is turned off into the helicopter turns into a fireball on the ground.
To find Austin 'Danger' Powers height before deploying the jetpack, use the same equation but leave out the distance and substitute in the 7.0s. During this time he is in free fall, so accelerating at 9.8 m/s/s
d = ut + 1/2at^2 = 66*7 - 4.9 *7^2 = 221.9 m.
Now, that was starting from a point already 363 m above the ground, and we did everything right with directions and it's a positive number, so his overall height above the ground is 363 + 221.9 = 584.9 m.
We need to know his velocity at this point as a starting point for the next part:
v = u + at = 66 - 9.8*7 = -2.6 m, so he's still travelling upward at this moment.
This is 7 s after the helicopter engine turned off, so 17.66-7 = 10.66 s before the helicopter hits the ground. So for 17 s, instead of falling at 9.8 m/s/s he falls at 2 m/s/s
d = ut + 1/2 at^2 = 17.66*10.66 - 1/2*2*10.66^2 = 74.65 m
This is still a positive number, so his height has increased yet again (due to the initial upward velocity)
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