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22

It will travel along a parabola (ignoring drag from the air here), initially with upward velocity, as you describe in your first scenario. You're correct that the only force acting on the box is its weight, but this means it will have downward acceleration immediately, not necessarily downward velocity. Eventually the downward acceleration will lead to ...


12

Your question is not weird, is legitimate. It is possible, it exists, can be of use and it is called jerk, jolt, surge or lurch, and is defined by any of the following equivalent expressions: $$\vec j(t)=\frac {\mathrm{d} \vec a(t)} {\mathrm{d}t}=\dot {\vec a}(t)=\frac {\mathrm{d}^2 \vec v(t)} {\mathrm{d}t^2}=\ddot{\vec v}(t)=\frac {\mathrm{d}^3 \vec r(t)} ...


4

Your question has great practical significance: it is the very essense of loft bombing/LABS. Since nuclear weapons can damage/destroy an attacking plane, it was deemed necessary to devise a method to release the bomb and to increase the separation between the air burst and the aircraft. I recommend watching this training video: ...


3

There are many reasons why acceleration would not be constant. Books often don't mention it because they are getting mathematicians used to the concept of first derivatives before moving onto second, third etc... Consider the following example which ends up leading to Tsiolkocsky's Rocket Equation: http://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation A ...


3

Your approach is correct, now simply add indices to everything, i.e. $$y_i = v_{0,i}t_i + \frac12 a_it_i^2\quad\text{where } i\in\{1,2\}$$ and note that $t_2 = t_1 - 2\,\text{s}$. Then solve $15\,\text{m}\stackrel!=y_1 - y_2$.


3

It depends on what direction you assign to be positive in your coordinate system. To avoid confusion, just remember which direction acceleration is acting and which direction you assigned to be positive.


3

I think your plotting routine is expecting angles in radians, not degrees. The starting angle of both graphs is wrong. The first is not $45^\circ$, and the second is nowhere near $30^\circ$. $\tan(30)=-6.4$ (if $30$ is in radians) and that would explain the negative start. Also: did you square the cosine in the denominator? The formulas above the graph ...


3

There are two parts to this problem. In the first part, the stone is rising with the balloon. It has a certain acceleration for a certain time. At the end of that it will have reached a certain height and velocity. Calculate it. Then the stone is released. With the initial velocity and height calculated above, it now starts dropping. It is now subject to ...


2

As an another attempt, I calculated the coefficients of the cubic regressions that describe the NIST data. I first calculated the cubic regression coefficients as a function of pressure for viscosity as a function of temperature. In other words, I calculated the coefficients for each isobar. In equation form that is, $$\mu ...


2

I do not know if it is "plausible" (I do not think so), however a trivial model can be constructed for the one-dimensional case with continuous forces depending on velocities, for $c>0$ constant: $$F_{12}(v_1,v_2) = c\sqrt{|v_1-v_2|} \quad\mbox{and}\quad F_{21}(v_1,v_2) = -c\sqrt{|v_1-v_2|}$$ The system of these two particles does not admit a unique ...


2

This does exist, and it is called Jerk, see the Wikipedia page on jerk. It is used quite frequently in physics concerning humans, as we are able to sense this, and there are limits to how much jerk a human can endure. It is, however, quite abstract and therefore more difficult to comprehend, which might be the reason that lower level textbooks do not ...


2

When you run into a problem like this, there are several things to consider: Systematic error. For example, when you say "three, two, one, go" you never release exactly on "go" (plus, there is a time delay between when the person says "go" and when you hear it...). Other example: the floor is not level (as suggested by ACuriousMind). When you measure the ...


2

$v$ has the units of $L\cdot{T^{-1}}$ $a$ has the units of $L\cdot{T^{-2}}$ $x$ has the units of $L$ $(L\cdot{T^{-1}})^2=(L\cdot{T^{-2}})\cdot{}L^P$ $L^2\cdot{T^{-2}}=L^{(P+1)}\cdot{T^{-2}}$ Comparing terms, $2 = P + 1 => P = 1$


1

The accepted answer is (subtly) wrong. While the projectile will indeed initially have an upward velocity, the shape it traces out is that of an ellipse, not a parabola -- remember, the earth is not flat.


1

If you impact the second body its axis of percussion it will purely rotate. By carefully choosing the inertial properties of the two objects you can make the first object stop translating in the process. See this post for more details on a particle to rod impact.


1

Use the standard equation: $$s=\frac{v^2-u^2}{2a}$$ break the problem into two halves: one accelerating and one decelerating. Taking the sum of both equations you should get the total distance traveled as a sum of some terms, all of which you know except the speed in the middle which should cancel out (assuming magnitude of $a$ is equal in both cases). ...


1

Use $d=v_i t +\frac{1}{2}a t^2$ and then substitute from the first equation that $v_i=-a t$. You get $$d= -at^2+\frac{1}{2}at^2=-\frac{1}{2}at^2$$ from which you can infer $$ a=-\frac{2d}{t^2}$$ Once you have the acceleration you get the initial velocity from $v_i=-a t$.


1

If you take the time of throwing of the first object as $0$ then the second object will start falling at $0+2$ second.Now for the second one take the time of its start of fall as 0 and so the ending time will be 2 sec less i.e$0+2 \longrightarrow 0$ and$t \longrightarrow t-2$ So for the second one $$ \frac {dx} {dt} =u+at$$ $$ dx=udt +atdt$$ $$\int^y_0 ...


1

In the first picture, the initial velocity in x would be $v_i\cos(a)$, and in the second picture, the initial velocity in x would be $v_j\cos(0)$ If you want to use the "textbook theory" (second picture), you cannot assume the red line in the second picture is the same as the red line in the first picture. It is, however, equivalent to the peak of the curve ...


1

If your position is in 3D space (which means your position vector must be defined), then there is no distinction between displacement and change in position. $s=\boldsymbol{R_f-R_i}=\Delta\boldsymbol{R}$ , where $s$ is displacement and $R$ is position. However, in $v = ds/dt$, $ds$ does not mean change in displacement but rather an infinitesimally small ...


1

As you propose let's find a "trick" and take advantage of the relative speed of the vehicles. Let's find out when they meet first. Let's imagine we are sitting in the slower vehicle. We say we are "in a frame of reference attached to the slower vehicle". From our perspective the faster vehicle is moving towards us at 50-25 = 25km/h. Still from our ...


1

Yes, you defined the zero of time as when the particle is at $(4,0,0)$. It passed through $(1,-1,5)$ one second before that.



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