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14

On the whole, static friction is higher than dynamic friction. This means that if you can brake without your wheels skidding, you will come to a halt more quickly. So let's assume that the truck brakes without skidding, and see where that gets us. Let's assume that your truck has weight $W = Mg$ with a haystack with additional weight $w = mg$ on top. ...


14

Not to detract from Floris' answer, but I think this is an instance where it is nice to think in terms of limits. If the hay is tied down, you're stopping an object with mass (truck + hay). If the hay isn't tied down, but on a sufficiently sticky surface such that it doesn't move, it should be the same as stopping it if it were fixed, since the outcome is ...


14

There are several interesting things going on when a car turns. First - let's take the simple diagram of two front wheels turned by 45°: As you can see, the top tire would like the car to turn around the point $C_1$, but the bottom tire (at the same angle) wants to turn around $C_2$. This means that in reality both tires will experience some lateral ...


7

Correct me if I'm wrong, I believe that from a paintball gun when shooting there is no smoke, vapour or other visible exhaust and not even recoil. No visible signal. The dodger must rely on sound. Sound reaching the ears. The speed of sound in air is around $v=330\:\mathrm{m/s}$. Sound will propagate the $x=8\:\mathrm{m}$ and reach the dodger in ...


6

Integrate the jerk 3 times then using starting conditions to work out the integration constants.


4

Inertia is an intrinsic characteristic of the object related to its mass. Inertia tells you how much force it will take to cause a particular acceleration on the object. Momentum is a function of an object's mass and velocity. Momentum is a measure of the kinetic energy of the object. A massive object can have any momentum (at least as long as its velocity ...


4

Momentum: The resistance of an object to a change in its state of motion. That sounds like a fishy definition of momentum to me. A slightly better definition, at least at your level, is that momentum represents the "amount of motion" an object has. Granted, "amount of motion" is a very vague term, but it stands to reason that if "amount of motion" were ...


3

No. Automobile tires to not expand radially to any great extent - the steel belts will keep that from happening. So, the tire radius still determines how far the car travels per rotation. Now, if your tires are slipping on the road, or are slipping with respect to the rims, than yes you have speedometer problems, but you have lots of other problems as ...


3

The angular acceleration $\alpha$ is just $\alpha = a/r = 1/200\textrm{rad/s}^2$ where $a$ is the tangential acceleration and $r$ the radius of the circle. Here, we can see that $\alpha$ is constant, which allows us to use the constant acceleration equations in their angular form. For (b) you can use the angular constant acceleration equations. You have ...


3

This is the simplest analogy I could think of. Imagine a long narrow carpet sliding across a huge ice rink at 1kph. On the rear end of the carpet stands a very fat (200kg) man wearing roller skates. You want to bring him to a standstill. You could grab the man and dig your ice skates into the ice until he eventually stops. Alternatively, you could grab the ...


3

You should always apply energy conservation, and it ought to hold in all reference frames, including the frame in which the sigma is at rest. In the sigma's rest-frame, $$ E_{\text{initial}} = E_\Sigma = m_\Sigma $$ and $$ E_{\text{final}} = E_\Lambda + E_\pi \ge m_\Lambda + m_\pi $$ Thus we have that, $$ E_{\text{initial}} < E_{\text{final}} $$ The ...


3

You are using the word "linear" in two different ways. When an object moves along a straight line we can say its motion is linear - but that does not mean its acceleration is zero. Just that the acceleration points along the same direction as the velocity (so no change in the direction of the motion). The second meaning of "linear" is in the exponents of ...


2

In reality you have $a(v) = g - \beta v$ and you want $v(t)$ and $x(t)$. This can be achived with direct integration, instead of coefficient matching and diff. equs. $$ t = \int\limits_0^v \frac{1}{a(v)}\,\mathrm{d}v= \int\limits_0^v \frac{1}{g-\beta v}\,\mathrm{d}v = -\frac{1}{\beta} \ln \left(1- \frac{\beta v}{g} \right)$$ or $$ v(t) = \frac{g}{\beta} ...


2

Definition As in the question we'll define our bicycle as the location in two dimensions of our contact patches $r_1$ and $r_2$ for the rear wheel and front wheel respectively. All other aspects of the bicycle will be ignored. Assumptions The first assumption corresponds to the distance between the wheels being constant: $$|r_2-r_1|=C'$$ ...


2

I don't understand why this question has not been answered in the comments. If there's no friction, then the "direction of the wheels" cannot turn the car. It's as though the car were on a sheer-ice skidpad. (Those of us who have driven in winter weather know the feeling of encountering a patch of "black ice".)


2

Arrows and paintballs go about the same speed, 200-300 feet per second. In episode 109, Mythbusters confirmed a professional ninja can block an arrow with a sword. It's plausible. However, distance is going to make a difference. I don't have the episode to check how far away the archer was, but in your question 24 feet (8 meters) is really close. It ...


2

The first one is basically derived from the velocity-time graph. As the acceleration is constant the area under the graph represents displacement (s). Thus by the formula to get the area under a trapezium. $$\text{Area} =\frac{1}{2} ( \text{sum of the parallel sides}) \times ( \text{vertical height})$$. We have $$S = \frac{1}{2} ( v + u)\cdot t$$.


2

That is partially correct. You are missing a critical part of Newton's 2nd law: $F_{net} = m a$. It's not just any one force that equals the mass times the acceleration. It's the "net force" - the vector sum of all the forces. So a car that is traveling at a constant speed may have many forces on it (gravity, the normal force, friction, air resistance, ...


2

The time that the person has to dodge is $ t=\frac{8 m}{160 m/s}=0.05 s $ So he has to move at a minimum speed of $ v=\frac{0.50 m}{0.05 s}=10m/s $


2

For any isolated object, i.e. one that is having an external force applied to it, the frame that object lives in is locally inertial. By locally I mean that if you consider a small enough region of spacetime around the object it will be impossible to tell the frame isn't inertial. How small this volume is depends on how curved spacetime is. This is an ...


2

Knowing only "jerk" (third derivative of position), you cannot determine the distance traveled. To get distance traveled (or equivalently, position as a function of time) from jerk, you need to integrate three times. Each integration produces a constant of integration representing an initial value; your final equation looks something like this: $$p(t) = ...


2

Starting out with the general $$z := h + v~t - \frac {g}{2}~t^2,$$ where $t \ge 0$ is the duration since the stone had been released, $z[~t~]$ is the remaining "height above ground" of the stone, $h$ is the "heigt above ground" of ballon and stone at the release, and $v$ is the vertical speed of the ballon at the release. Given $z[~\frac{2~u}{g}~] := ...


2

Your error is that you assume it takes 2.00s seconds for the ping to reach the cliff and an additional 1.98s to return. Without knowing the distance to the cliff, we can't qualify that assumption. Besides that, the only thing we need for the calculation is the difference in period of the outgoing and incoming pings. With a single ping you can calculate the ...


2

The round trip time of the ping is unknown; but we do know that the difference in round trip time between sub stationary and sub moving is 0.02 seconds. Let us write $D$ for the distance to the cliff when you send the ping; if you are traveling at a speed $v$, and the speed of sound in water is $c$, then we can write down the round trip time as follows ...


2

...a truck in motion and it has stack of hay (lets suppose) on the back. Now if the truck comes to a sudden stop will it stop faster if the force exerted by the truck on hay had overcome the friction force (another wording: will it be faster if the hay slips forward) or will it stop faster if the hay remains constant. I tried to find a braking ...


1

What a strange question. Is $x(t)$ the distance measured along the track? Without defining $x$, there is nothing else to say. And if it is the distance, then it's the distance. Along the track. And the car is traveling along the track. So x is the distance. Which is what is asked. I suggest you read ...


1

I am going to make some general comments, expecting that they will lead for further clarification of the question. First - when a boat sails into the wind (as the one in this question does), it is important to realize that (all these are idealized statements... real sailing is a lot more complicated): a) the force of the wind is approximately normal to the ...


1

This can be understood in terms of vector differentiation and the dot product. Take the example that $v \perp r$. The change in the square of the displacement is $$\frac{d}{dt}r^2$$ $$=2r \cdot v$$, and if they are perpendicular, the dot product is zero.


1

the areas of two rectangles are equal, and the sides are proportional: LHS rectangle with sides: $(v+u)$ and $(v-u)$, -- from $(v^2-u^2)$ RHS rectangle with sides: $\sqrt{2}$ boosted by acceleration and displacement: $(a)$ and $(s)$ for simplicity assume $v>u$ and displacement $s$ as dependent on acceleration $a$, rearranging ...


1

Suppose you have a constant acceleration $a$. Acceleration is the derivative of velocity with respect to time, so we have: $$ \frac{dv}{dt} = a $$ and if we integrate this we get: $$ v = at + C $$ where $C$ is the constant of integration. To find $C$ we note that when $t = 0$ the velocity is equal to $u$ (the initial velocity) so that means $C = u$ and ...



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