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0

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.


0

If you plug in numbers during the solution it is a good idea to carry units: you would-be have seen your mistake immediately. Starting off 500mph is about 220 m/s (please correct me if I am wrong). We divide this by t= 1 s and get average acceleration of $220 m/s^2$. To express this answer in terms of g = 9.8m/s^2, we divide: This average acceleration ...


0

I would not worry about air resistance for the balloon - being large and filled with water it will matter less than for the little ball. I would, however, worry about the strength of the catapult - how does its launch velocity change with mass of the object. If the catapult is made with an elastic that is pulled back by e certain amount then it van do a ...


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 ...


0

For the vertical motion you can ignore horizontal initial velocity of one bullet. This horizontal initial velocity just stays constant when air resistance is ignored. Therefore, in the vertical direction the two bullets undergo the same accelerated motion due to the gravitational force and will hit the bottom at the same time.


-3

You need to assume they both have the same initial velocities for anything interesting to happen. If they both have the same initial velocities, for instance both are fired from the same gun with the same powder load, removing air resistance. If you shoot one at the ground and shoot the other horizontally they will both be accelerated towards earth at ...


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

Whether the upward direction will be taken as positive & the downward direction as negative or vice versa simply depends upon you. It does not bother any physics. Generally, the downward direction is taken as negative. The negative sign indicates the opposite direction only of what you have assigned the positive direction. In Principles of Physics by ...


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.


0

Well, things don't fall up, so gravity would be negative. The longer version: For classical mechanics, the direction of a coordinate system is often arbitrary. For free fall problems I generally choose down to be positive (especially when I'm not dealing with an initial velocity, but it's a matter of preference). What you do need to know, no matter what ...


0

If up is positive, and gravity points down, then $a$ (acceleration due to gravity) would be downwards, so it will have a negative magnitude.


0

How much is the force of gravity on the $3m$ mass, and how much mass must it accelerate? Similar question for the $2m$ mass? Then carry on as you have...


0

Assuming you are considering motion with constant acceleration $a$ and initial velocity $v_0$, you know: $$s=v_0 t+\frac{1}{2}a t^2$$ You can solve that quadratic equation for t: $$t=-\frac{v_0}{a}+\sqrt{\frac{v_0^2}{a^2}+\frac{2s}{a}}$$ Your final velocity is then $$v=v_0+at=\sqrt{v_0^2+2as}$$. In the case that the motion started from rest: $$v=\sqrt{2as}$$ ...


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 ...


0

The equations $$y=-gt^2/2$$ and $$ t=\sqrt{\frac{2y}{-g}}$$ are equivalent expressions. The top equation is your typical freefall equation as you would see it in a physics textbook. However, since you are solving for the amount of time that your projectile is falling down, the textbook author rearranged the equation to solve for time. Unlike the comment ...


0

Circular motion is a motion with acceleration, because the velocity vector is changed during the motion. Therefore the engine must be on all the time, and the resulting force is changing the rockets velocity vector's direction, but not size (under other circumstances than those give in the question, the rocket could be using this same propulsive force to ...


0

It is undergoing uniform circular motion, so its linear velocity is tangential to the circle, and its acceleration is towards the centre of the circle. You can easily prove that it is tangential using calculus on complex numbers; Let $z=Re^{i \theta}=x+iy$ Differentiating both sides with respect to time, $\frac{dz}{dt}=Ri e^{i \theta}*\omega=i \omega z$, ...


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 ...


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 ...


0

Two tricks you can use. First - just solve the equations of motion. You can write down the position as a function of time: $$x(t) = x(0) + v\cdot t$$ where $x(t)$ is the position at time $t$, and $v$ is the velocity. Two equations, solve for $t$. Simpler still - look at the difference in velocity. If one goes at 25 km/h and the other goes at 50 km/h, the ...


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

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 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 ...


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 $15\,\text{m}\stackrel!=y_1 - y_2$.


0

Another equation of motion problem,very easy. $$v^2 = u^2 + 2.a.s$$ where $$v = \textbf{final velocity} $$ , $$u = \textbf{initial velocity}$$ , $$a=\textbf{ acceleration or in this case negative acceleration}$$ , $$s = displacement$$ . This equation is time independent. Now, to find $s$ , put the values: $$s = \frac{-1000^2}{2.(-10)} \implies s = ...


0

x is position $x$ $[L]$ v is velocity $\frac{\delta x}{\delta t}$ $[L][T]^{-1}$ a is acceleration $\frac{\delta v}{\delta t}$ $[L][T]^{-2}$ forget the constant factors ( 2 has no dimension ) from $v^2 = a \cdot x^P$ we get $[L]^2[T]^{-2}=[L][T]^{-2}[L]^{P}$ or $[L]^2=[L]^1[L]^{P}$ then $[L]^2=[L]^{1+P}$ valid when $P=1$ then the valid expression is ...


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, $2 = P + 1 => P = 1$


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 ...


0

Its pretty simplistic, if you know basic physics you will be aware that $\vec{A} = \frac{\Delta{v}}{t}$ where $\Delta{v}$ is the change in velocity which can be calculated by $v_{final} - v_{inital}$ where $v_{final}$ is the final velocity and $v_{inital}$ is the original velocity prior to acceleration and and also $t$ is the change in time for the velocity ...


0

The gradient of the line tells us the rate of change of velocity, which is acceleration. Therefore on an A-T graph, it would be a horizontal line, with a = 4.


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.


0

You're missing perspective in your question. With respect to the Earth, the box will travel in a parabolic arc, just as if you've thrown the box into the air with the same initial velocity. While the acceleration is certainly downwards after you drop it, its velocity is still upwards (and forward) until gravity changes that (and gravity is quite weak, so it ...


0

All objects that are in free fall (which is to say, no other forces are acting on said objects) will have a height above ground that can be predicted as such, until said objects reach the ground: h=-(g/2)(t^2) + vt + c where: g represents the acceleration due to gravity. On Earth, at sea level, this value is approximately 9.80665 m/s^2. t represents the ...


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: ...


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 ...


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 ...


-1

Someone please correct me if I'm wrong, I'm new.. The balloon is rising from the groung, so if you think about it, you have to calculate the speed the balloon reaches after 8 seconds and take it as the initial speed of the stone, then since we assume for simplicity that the air resistance is not significant, there is no other force acting on the stone, just ...


0

Because the initial velocity of the stone is not zero.


0

You said this is a one dimensional problem, and the drag force is proportional to the square of the velocity $v$, so $$ F = - b v^2 = m \frac{dv}{dt} \ \ \ \Rightarrow \ \ \ \int_{v\left(0\right)}^{v\left(t\right)}\frac{dv}{v^2} = - \frac{b}{m} \int_0^t dt' $$ The above integral (I'll let you do it, or see this) gives $v\left(t\right)$. The average velocity ...


0

You need to use a differential equation. We can write Newton's second law as $$ F = m a = m \frac{dv}{dt} = m \frac{d^2x}{dt^2} $$ Your wind resitance will look like $$ F = - c v = - c \frac{dx}{dt} $$ so you can write $$ m \frac{d^2x}{dt^2} = - c \frac{dx}{dt} $$ And it's possible to solve for x. Once you have x, the average speed is just x(t)/t ...


0

I think the basic problem here is that a magnetic gradient is required for levitation, whereas the Earth's magnetic field is very close to uniform on the scale of any human-scale crafts. Try using the right-hand rule to figure out the direction of the magnetic force from a uniform magnetic field at two points on opposite sides of a circular current loop ...


-2

Answer for a: The second ball is released from rest and assumes going down is negative: u=0 g=-9.81 v=? h=-20 t=? By using the kinematics equation substitute the value in following equation: $$ s=ut^2+\frac12 at^2$$ So: $$-20=-\frac12 9.81 t^2 $$ Therefore: t=2.02 s. It said it released after 1 s so: t=3.02 s. The first ball is projected vertically ...


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

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.


0

Like New_new_newbie said, the question asks you to find the distance travelled in its third second (from t=2 to t=3 seconds). So just find x(2) and x(3), and the difference should give you 29m.


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 ...


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 ...


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)} ...


0

You are almost done. If you simplify the express you found for $v_{avg}$ and compare it with the initial equation for $v_{avg}$, in which you have to substitute $v_i$ for $v(t_i)$ and $v_f$ for $v(t_f)$, you will see that they are equivalent. Another way of deriving this is by expressing $v(t)$ in terms of $v_i$ and $v_f$: $$ v(t) = v_i + \frac{v_f - ...


1

There is one error in the derivation, if you want to have $v(t_i)=v_0$, you must have $$v(t) = v_0 + a(t-t_i)$$ You also have to use the fact that $v_f = v(t_f)$. Once you use all this, you should be able to divide out $t_f-t_i$ in the corrected version of your last line and get the result you seek.



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