Imagine the runway as a slope of fading push: the jet begins at rest and the acceleration shown in the graph decreases linearly from 22.5 meters per second squared at the start to zero at 150 meters. Before diving into integrals, ask why we switch from time to space: when acceleration depends on position, following the plane along the runway — meter by meter — gives a direct way to find speed after a given distance.
Storyframe and analogy: think of acceleration like water pressure that changes along a hose. The speed the plane reaches after sixty meters is like the volume of water delivered after a certain length of hose — you care about how pressure varies with distance. The kinematic link that does this is v times dv/ds equals a of s, which turns the problem into an integral over distance.
From the graph we read a of s equals 22.5 minus 0.15 s. Integrate v dv from zero to v and a of s ds from zero to sixty. The left side becomes one half v squared. The right side evaluates to 22.5 times 60 minus one half times 0.15 times 60 squared, which is 1080. So one half v squared equals 1080, giving v squared equals 2160, and v equals 12 times the square root of 15 meters per second, about 46.5 meters per second.
Why it matters: the early strong acceleration deposits most of the kinetic energy; even as acceleration drops off, that momentum carries the plane forward. Changing your viewpoint from time to space often turns a messy problem into a clear path.
A thought to close on: the most useful perspective is often the one that follows the object itself — change the frame, and the solution reveals itself.
Kinematics pursuit problem (Irodov): A moves uniformly with speed v, always directed toward B. B moves in a straight line with speed u (u < v). Initially v ⟂ u and the separation is l. How long until they meet? Try solving before checking the solution segment.
10X Physics
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10X Physics
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10X Physics
Imagine the runway as a slope of fading push: the jet begins at rest and
the acceleration shown in the graph decreases linearly from 22.5 meters per second
squared at the start to zero at 150 meters. Before diving into integrals, ask
why we switch from time to space: when acceleration depends on position, following
the plane along the runway — meter by meter — gives a direct way to find speed
after a given distance.
Storyframe and analogy: think of acceleration like water pressure that changes
along a hose. The speed the plane reaches after sixty meters is like the volume
of water delivered after a certain length of hose — you care about how pressure
varies with distance. The kinematic link that does this is v times dv/ds equals
a of s, which turns the problem into an integral over distance.
From the graph we read a of s equals 22.5 minus 0.15 s. Integrate v dv from zero
to v and a of s ds from zero to sixty. The left side becomes one half v squared.
The right side evaluates to 22.5 times 60 minus one half times 0.15 times 60 squared,
which is 1080. So one half v squared equals 1080, giving v squared equals 2160,
and v equals 12 times the square root of 15 meters per second, about 46.5 meters
per second.
Why it matters: the early strong acceleration deposits most of the kinetic energy;
even as acceleration drops off, that momentum carries the plane forward. Changing
your viewpoint from time to space often turns a messy problem into a clear path.
A thought to close on: the most useful perspective is often the one that follows
the object itself — change the frame, and the solution reveals itself.
8 months ago | [YT] | 5
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10X Physics
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10X Physics
Kinematics pursuit problem (Irodov): A moves uniformly with speed v, always directed toward B. B moves in a straight line with speed u (u < v). Initially v ⟂ u and the separation is l. How long until they meet? Try solving before checking the solution segment.
8 months ago | [YT] | 4
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10X Physics
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