Answer:


Explanation:
Wave 1, 
Wave 2, 
Wave 3, 
Wave 4, 
The general equation of travelling wave is given by :

The value of
will remain the same if we take phase difference into account.
For first wave,



For second wave,



For the third wave,



For the fourth wave,



It is clear from above calculations that waves 1 and 3 have same time period. Also, wave 2 and 4 have same time period. Hence, this is the required solution.
Well, I guess you can come close, but you can't tell exactly.
It must be presumed that the seagull was flying through the air
when it "let fly" so to speak, so the jettisoned load of ballast
of which the bird unburdened itself had some initial horizontal
velocity.
That impact velocity of 98.5 m/s is actually the resultant of
the horizontal component ... unchanged since the package
was dispatched ... and the vertical component, which grew
all the way down in accordance with the behavior of gravity.
98.5 m/s = √ [ (horizontal component)² + (vertical component)² ].
The vertical component is easy; that's (9.8 m/s²) x (drop time).
Since we're looking for the altitude of launch, we can use the
formula for 'free-fall distance' as a function of acceleration and
time:
Height = (1/2) (acceleration) (time²) .
If the impact velocity were comprised solely of its vertical
component, then the solution to the problem would be a
piece-o-cake.
Time = (98.5 m/s) / (9.81 m/s²) = 10.04 seconds
whence
Height = (1/2) (9.81) (10.04)²
= (4.905 m/s²) x (100.8 sec²) = 494.43 meters.
As noted, this solution applies only if the gull were hovering with
no horizontal velocity, taking careful aim, and with malice in its
primitive brain, launching a remote attack on the rich American.
If the gull was flying at the time ... a reasonable assumption ... then
some part of the impact velocity was a horizontal component. That
implies that the vertical component is something less than 98.5 m/s,
and that the attack was launched from an altitude less than 494 m.
Answer:
In a time-position graph (s-t graph):
slope = velocity
In a time-velocity graph (v-t graph):
slope = acceleration
area under graph = change in displacement (distance travelled)
In a time-acceleration graph (a-t graph):
area under graph = change in velocity