-- We know that the y-component of acceleration is the derivative of the
y-component of velocity.
-- We know that the y-component of velocity is the derivative of the
y-component of position.
-- We're given the y-component of position as a function of time.
So, finding the velocity and acceleration is simply a matter of differentiating
the position function ... twice.
Now, the position function may look big and ugly in the picture. But with the
exception of 't' , everything else in the formula is constants, so we don't even
need any fancy processes of differentiation. The toughest part of this is going
to be trying to write it out, given the text-formatting capabilities of the wonderful
envelope-pushing website we're working on here.
From the picture . . . . . y (t) = (1/2) (a₀ - g) t² - (a₀ / 30t₀⁴ ) t⁶
First derivative . . . y' (t) = (a₀ - g) t - 6 (a₀ / 30t₀⁴ ) t⁵ = (a₀ - g) t - (a₀ / 5t₀⁴ ) t⁵
There's your velocity . . . /\ .
Second derivative . . . y'' (t) = (a₀ - g) - 5 (a₀ / 5t₀⁴ ) t⁴ = (a₀ - g) - (a₀ /t₀⁴ ) t⁴
and there's your acceleration . . . /\ .
That's the one you're supposed to graph.
a₀ is the acceleration due to the model rocket engine thrust
combined with the mass of the model rocket
'g' is the acceleration of gravity ... 9.8 m/s² or 32.2 ft/sec²
t₀ is how long the model rocket engine burns
Pick, or look up, some reasonable figures for a₀ and t₀
and you're in business.
The big name in model rocketry is Estes. Their website will give you
all the real numbers for thrust and burn-time of their engines, if you
want to follow it that far.
<u>Answer:</u>
For 1: The correct option is Option C.
For 3: The final velocity of the opponent is 1m/s
<u>Explanation: </u>
During collision, the energy and momentum remains conserved. The equation for the conservation of momentum follows:
...(1)
where,
are the mass, initial velocity and final velocity of first object
are the mass, initial velocity and final velocity of second object
<u>For 1:</u>
We are Given:
Putting values in equation 1, we get:
Hence, the correct answer is Option C.
Impulse is defined as the product of force applied on an object and time taken by the object.
Mathematically,
where,
F = force applied on the object
t = time taken
J = impulse on that object
Impulse depends only on the force and time taken by the object and not dependent on the surface which is stopping the object.
Hence, the impulse remains the same.
Let the speed in right direction be positive and left direction be negative.
We are Given:
Putting values in equation 1, we get:
Hence, the final velocity of the opponent is 1m/s and has moved backwards to its direction of the initial velocity.
Answer:
The time is 5.71 sec.
Explanation:
Given that,
Acceleration 
Initial velocity = 24.0 m/s
We need to calculate the time
Using equation of motion
v = u+at[/tex]
Where, v = final velocity
u = inital velocity
t = time
a = acceleration
Put the value into the formula
Hence, The time is 5.71 sec.
