Identify the velocity-versus-time plot(s) that correspond to motion under a constant, non-zero acceleration.
1 answer:
Answer:
See explanation
Explanation:
The question is incomplete because the images were not attached but I will try to help you as much as possible.
Constant acceleration implies that the velocity increases uniformly with time.
The graph of constant acceleration is a straight line graph having a slope. The slope of the graph is constant at any point along the straight line.
The image attached shows a velocity-time graph depicting constant acceleration.
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Answer:
a) t = 20 [s]
b) Can't land
Explanation:
To solve this problem we must use kinematics equations, it is of great importance to note that when the plane lands it slows down until it reaches rest, ie the final speed will be zero.
a)
where:
Vf = final velocity = 0
Vi = initial velocity = 100 [m/s]
a = desacceleration = 5 [m/s^2]
t = time [s]
Note: the negative sign of the equation means that the aircraft slows down as it stops.
0 = 100 - 5*t
5*t = 100
t = 20 [s]
b)
Now we can find the distance using the following kinematics equation.
x - xo = distance [m]
x -xo = (0*20) + (0.5*5*20^2)
x - xo = 1000 [m]
1000 [m] = 1 [km]
And the runaway is 0.8 [km], therefore the jetplane needs 1 [km] to land. So the jetpalne can't land
1) The variable that is different in the two cases is , the duration of the collision
2) The change in momentum is the same in the two cases
3) The impulse is the same in the two cases
4) Case B will experience a greater force
Explanation:
1)
The variable that is different in the two cases is , the duration of the collision.
In fact, in the first case the wall is padded: this means that the collision will be "softer" and therefore will last longer, so the duration of the collision, , will be larger.
In the second case instead, the wall is unpadded: this means that the collision is "harder" and so it will last less time, therefore the duration of the collision will be smaller.
2)
The change in momentum in the two cases is the same.
In fact, the change in momentum is given by:
where:
m is the mass of the student
u is the initial velocity
v is the final velocity
In both cases, we have:
m = 75 kg
u = 8 m/s
v = 0 (they both comes to rest)
Therefore, the change in momentum is
3)
The impulse in the two cases is the same.
In fact, impulse is defined as the product of force applied, F, and duration of the collision, :
However, the force can be rewritten as product of mass (m) and acceleration (a), according to Newton's second law:
So the impulse is
The acceleration can be rewritten as rate of change of velocity:
So the impulse becomes
So, the impulse is equal to the change in momentum: and since in the two cases the change in momentum is the same, the impulse is the same as well.
4)
The force in the collision is related to the impulse by
where
J is the impulse
F is the force
is the duration of the collision
The equation can be rewritten as
In the two situations described in the problem (A and B), we already said that the impulse is the same (because the change in momentum is the same). However, in case A (padded wall) the time is longer, while in case B (unpadded wall) the time
is shorter: since the force F is inversely proportional to the duration of the collision, this means that in case B the student will experience a greatest force compared to case A.
Learn more about impulse:
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