Answer: 88 m/s
Explanation:
If we are talking about an acceleration at a uniform rate, we are dealing with constant acceleration, hence we can use the following equation:
(1)
Where:
Is the final velocity of the plane (we know it is zero because we are told the pilot stops the plane at a specific distance)
Is the initial velocity of the plane
is the constant acceleration of the plane
is the distance at which the plane stops
Isolating from (1):
(2)
(3)
Finally:
This is the veocity the plane had before braking began
OPTION (C) IS CORRECT
ANSWER - UNIFORM MOTION
If there is no change in the velocity of the object then it is known to be in <u>UNIFORM MOTION</u>.
EXPLORE MORE:-
EXAMPLE - A CAR COVERS A DISTANCE OF 15 KM WE'D EVERY 2 HOURS
FOR AN UNIFORM MOTION, ACCELERATION IS ZERO , AS THERE IS NO CHANGE IN VELOCITY....
-THANKS.!!
Answer:
9.5 m/s
Explanation:
Distance, S = 150m
Acceleration, a = 0.3 m/s^2
Initial velocity, u = 0 m/s
Final velocity, v
Use kinematics equation
v^2 - u^2 = 2aS
v^2 - 0 = 2*0.3*150 = 90
v = sqrt(90) = 9.49 m/s
As the air becomes warmer, heat<span> is transferred </span>between<span> molecules and kinetic</span>energy<span> is created which produces </span>thermal energy<span>. As the molecules move faster to transfer </span>heat<span>, the </span>temperature<span> also increases.</span>
-- 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.
