If a rock falls from a height of 20 meters on the planet Jupiter, then its height (H) after T (t= seconds) is approx:
H(t)= 20-13t^2
What is the average velocity of the rock from t=0 to t=1
What is the instantaneous velocity at time t=1
What is the acceleration of the rock?
When does the rock hit the ground?
I have no idea when the damn rock hits the ground! lol. something to do with the acceleraion of gravity.
2007-10-29 03:51:56 · 8 answers · asked by girlygirl 1 in Science & Mathematics ➔ Mathematics
H(t)= 20-13t^2
velocity, V(t) = dH/dt = -26t m/s
acceleration, A(t) = dV/dt = -26 m/s^2
What is the average velocity of the rock from t=0 to t=1?
Ans: Since V(t) is a linear function,
Average Velocity = [V(0) + V(1)] / 2 = -13 m/s
What is the instantaneous velocity at time t=1?
Ans: V(1) = -26 m/s
-ve value indicates that it is travelling towards the ground
What is the acceleration of the rock?
Ans: acceleration, A(t) = -26 m/s^2
When does the rock hit the ground?
Ans: It hits the ground when
H(t)=0
==> 20-13t^2 = 0
==> t = (20/13)^(1/2)
==> t = 1.24 sec
~~~
.
2007-10-29 04:05:26 · answer #1 · answered by analog 2 · 0⤊ 0⤋
H(t)= 20-13t^2
velocity found be differentiating,
H'(t) = V(t) = -26t
at t=0, V (0) = 0, at t=1, V(1) = -26,
(a) so average speed is 26/2 = 13m/s
(b) V(1) = -26 m/s
(c) acceleration is -26m/s^2, since V'(t) = -26
(d) Rock hits ground when, H = 0, i.e. when
0 = 20 - 13t^2, i.e. when, t = +-20/13 seconds, and we are looking at positive time, so, t = 20/13 seconds
2007-10-29 04:04:35 · answer #2 · answered by tsunamijon 4 · 0⤊ 0⤋
Distance covered in 1 s
= [20 - 13(0)^2] - [20 - 13(1)^2) = 13 m
Av. velocity from t = 0 to t = 1 is
13 m/s in downward direction.
instantaneous velocity = dH/dt = -26t
=> (dH/dt) at (t = 1) = -26 m/s upwards
=> 26 m/s downwards
Acceleration = d2H/dt2 = -26 in the upward direction
=> 26 m/s^2 downwards
H = 0 when the rock hits the ground.
0 = 20 - 13t^2
=> t = √(20/13) = 1.24 s.
2007-10-29 04:05:24 · answer #3 · answered by Madhukar 7 · 0⤊ 0⤋
H'(t) = -26t
What is the average velocity of the rock from t=0 to t=1?
[H'(1) - H'(0)]/2 = -13 m/s
What is the instantaneous velocity at time t=1?
H'(1) = -26 m/s
What is the acceleration of the rock?
H''(t) = -26 m/s^2
When does the rock hit the ground?
H(t) = 0
t = sqrt(20/13) = 1.24 s
2007-10-29 03:59:11 · answer #4 · answered by sahsjing 7 · 0⤊ 0⤋
>> I have no idea when the damn rock hits the ground.
When the height is exactly zero is when the damn rock hits the ground. Set H(t) = 0 and solve for t.
2007-10-29 03:57:30 · answer #5 · answered by fcas80 7 · 0⤊ 0⤋
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2016-10-14 08:01:57 · answer #6 · answered by Anonymous · 0⤊ 0⤋
a. V = H' = -26t
...V(0) = 0, V(1) = -26, Vavg = -13 m/sec
b. V(1) = -26 m/sec
c. a = V' = -26 m/sec²
d. x = -1/2at²
...t = sqrt(-2x/a) = sqrt(-2*20/-26) = 1.24 sec
2007-10-29 04:00:35 · answer #7 · answered by gebobs 6 · 0⤊ 0⤋
I don't think jupiter even has a surface. It is a gas giant.
2007-10-29 04:01:39 · answer #8 · answered by minorchord2000 6 · 0⤊ 0⤋