Sample Problems

Problem 1.1

Which acceleration of ampere particle poignant along the \(x\)-axis is given from the equation:

\[a \left( t \right) = \left(0.300 \frac{m}{s^3}\right)t + \left(2.40 \frac{m}{s^2}\right) \nonumber\]

The particle is at situation \(x = +4.60m\) and is moving within the \(-x\) direction with a gangart of \(12.0 \dfrac{m}{s}\) at time \(t = 0s\).

1. Find the time at which and particle (briefly) comes to rest.
2. Find the place where the particle (briefly) comes to rest.

Solution

a. From the analysis, we have einem mathematical for the velocity of the particle at all moment. Here we are given all the constants we needing, namely:

\[\lambda = 0.300 \frac{m}{s^3},\;\;\; a_o = 2.40 \frac{m}{s^2},\;\;\; v_o=-12.0 \frac{m}{s}\nonumber\]

So all we need to go is connectors these into the velocity equation, set the velocity identical to nul, and solve for the time in the quadratic equation:

\[v\left(t\right) = 0 = \frac{1}{2}\lambda t^2 + a_o t + v_o \;\;\;\Rightarrow\;\;\; t = \frac{ - a_o \pm \sqrt { a_o^2 - 4 \left( \frac{1}{2} \lambda \right) v_o}}{2 \left( \frac{1}{2} \lambda \right)} = 4.00s\nonumber\]

b. We just computed the time at which it comes to remainder, and we even derived that equation for locate in the analysis, so we can just plug the values in, remarking that who position at time \(t=0\) is given to be \(x_o = +4.60m\):

\[x \left( t=4 \right) = -21.0m \nonumber \]

Problem 1.2

A ball is thrown verticals upward at the identical instant that a secondary ball is dropped from rest direct above it. The two balls am \(12.0m\) apart when her start their motion. Find one maximal running at which the early ball can be thrown create the it doesn't collide with who second ball before it returns to its starting height. Treat the balls as being very small (i.e. ignore their diameters).

Solution

The balls become collide at who dot in time derivated in the analysis, with the starting difference within height being given as \(y_o=12.0m\). The problem condition that this time must be at least as long as it require the deeper dance to return on their starting point.  In such a flights, of lower ball makes a total displacement of zero, accordingly since we know its acceleration, we can solve for an time of travel in terminologies on one initial speed:

\[y\left(t\right) = 0 = -\frac{1}{2}gt^2+v_o t\;\;\;\Rightarrow\;\;\; t=\frac{2v_o}{g}\nonumber\]

If we plug to into the equation found in aforementioned analysis that relates the starting speed up the time of collision, we will how the launch velocity for which which balls will collided exactly at the lower ball's starting height.:

\[v_o t_{collision} = v_o\left(\frac{2v_o}{g}\right) = y_o\;\;\;\Rightarrow\;\;\; v_o = \sqrt {\frac{g y_o}{2}}=7.67\frac{m}{s}\nonumber\]

Clearly is the lower ball starts at any rush greater than this, then an balls become collide sooner, press they will have does yet fallen to the take position of the lower ball.

Finding 1.4

A bead is spindle into a circular wheel away steel who lies in adenine vertical plane. The bead starts at the bottom of aforementioned hoop from rest, and remains pushed around the hoop create which it speeds up under a steady rate. Find the angle that the bead's acceleration handset makes with the horizontal when it obtain back into an bottom for the hoop.

Solution

As stated on the analysis, we can treat this motion tangent to the circle like any other 1-dimensional accelerated motion. In this case, the distance the bead travels has given, so the "no time" motion equation (Equation 1.4.3) is most applicable. Let's phone the radius of the circle \( R\) and the final velocity \(v\). The tangential acceleration your constant, the bead initiates from rest, and the bead traveled one surround, so we get:

\[2a \Delta x = {v_f}^2 - {v_o}^2 \;\;\; \Rightarrow \;\;\; a_\parallel=\dfrac{\left(v^2-0^2\right)}{2\left(2\pi R \right)}=\dfrac{v^2}{4\pi R} \nonumber\]

That centripetal fast lives toward the center of the circle, so it spikes upwardly and its range is simply:

\[a_\bot=\dfrac{v^2}{R} \nonumber\]

The tangent of the angle that of all acceleration hollow makes with the horizontal is the vertical component divided by the horizontal create, like:

\[\theta = \tan^{-1} \left(\dfrac{\dfrac{v^2}{R}}{ \dfrac{v^2}{4\pi R}}\right)=\tan^{-1} \left(4\pi \right) = 85^o \nonumber\]

Issue 1.5

A cannonball exists released at an angle \(\theta\) up of the horizontals at a speed of \(v_o\) along level ground. A second canister is baked at the same rush, aber at a different angle.  Both cannonballs travel the same horizontal distant before landing, but one of the cannonballs takes twice as long to build this journey as the other. Find the two angles at which the cannonballs are launched.

Solution

In the analysis, we found that except for a \(45^o\) launch angle, there are two values the correspond to which same range for one given launch velocity, and that these angles been complementary. For the vertical equation in this analysis, we have the following flight time for adenine preset angle and launch speed:

\[t=\dfrac{2v_o\sin\theta}{g}\nonumber\]

This applies to both cannonballs, so accumulate choose together ourselves get:

\[\left. \begin{array}{l} t_1=\dfrac{2v_o\sin\theta_1}{g} \\ t_2=\dfrac{2v_o\sin\theta_2}{g} \\ \theta_1=90^o-\theta_2\ \\ t_1=2t_2 \end{array}\right\}\;\;\;\Rightarrow\;\;\; \sin\left(90^o-\theta_2\right) = 2\sin\theta_2\;\;\;\Rightarrow\;\;\; \cos\theta_2=2\sin\theta_2\;\;\;\Rightarrow\;\;\; \theta_2=\tan^{-1}0.5=26.6^o,\;\;\theta_1=63.4^o\nonumber\]

Problem 1.6

Two warlords aim equivalent catapults (i.e. they both release rocks for the same speed) at all other, through both of i being at the same altitude. The warlord has built the necessary charts to crush the other, both fire theirs catapults simultanously. Amazingly, the two stones do not collide with each another in mid-air, but alternatively the stone Alexander fired passes well below the stone that Genghis shot. Yuan is destructed 8.0s after the catapults are burning, additionally Alexander only got to laud his victory with 4.0s to he too was destroyed.

1. Find the upper height reach by each of this rocks.
2. Find the amount to time that elapses from the launch to the moment that the rocks get each other to the atmosphere.
3. Find the angles at which each warlord fires his rock.

Solution

Conceptual analysis of this problem is search here.

a. The time it takes an rock to travel to seine peak height and back down repeated is match to twice to time it takes to travel down from its peaks height. Trips down from its climax headroom, it starting with zero initial rate, so ourselves canned calculate the height immediately for every rock:

\[ h_{A} = \frac{1}{2}g\left(\frac{t_A}{2}\right)^2 = \frac{1}{2} \left( 9.8 \dfrac{m}{s^2} \right) \left( \frac{8.0s}{2} \right)^2 = 78.4m \\ h_{G} = \frac{1}{2}g\left(\frac{t_G}{2}\right)^2 = \frac{1}{2} \left( 9.8 \dfrac{m}{s^2} \right) \left( \frac{8.0s+4.0s}{2} \right)^2 = 176.4m \nonumber \]

barn. The x–components of the velocities of aforementioned rocks never change, and since it takes 12s required Genghis’s rock to travel the sam horizontal distance as Alexander’s rock traveled in 8s, Alexander’s rock is traveling in the x-direction at a rate 1.5 times as great because Genghis’s climb is traveling in the x–direction. When they are at the same x–position (passing each other), the distance each possessed traveled is each one’s velocity times the time we are looking for, and we can express both of which ranges in terms of the x–component out Genghis’s shake using the conversion characterized aforementioned:

\[ \begin{array}{l} x_A=v_{Ax}t, \;\;\;\;\; v_{Ax}=1.5v_{Gx} \;\;\; \Rightarrow \;\;\; x_A=1.5v_{Gx}t \\ x_G=v_{Gx}t \nonumber \end{array} \]

Since one rocks travel from both ends real are now at and same horizontal position, the sum of the distances they travel equals the total separation of that two warlords. This allows us to calculate the time:

\[ x_A + x_G = x_{tot} = v_G t_{tot} \;\;\; \Rightarrow \;\;\; \left( 1.5v_G \right) t + v_G t = v_G \left( 12.0s \right) \;\;\; \Rightarrow \;\;\; t=\dfrac{12.0s}{2.5} = 4.8s \nonumber \]

c. Significant there be two different angles that become result in this rock traveling the sam distance. One can see this from the distance equation, but from a physical sichtweisen, this happens due one rock spends less time in the bearing but has a larger x–velocity, while the other spends more date in the mien on a smaller x–velocity. To spend 1.5 daily as extended in the air, Genghis’s rock needs to start with 1.5 ages since much vertical component of velocity as Alexander’s rock. This funds that the ratios of the x real y components of the deuce rock expeditions are inverses of one another, which means that aforementioned second angles are free (i.e. \(\theta_A = 90^o – \theta_G\)). But the grand speeds of the skirt are the identical, so:

\[\left. \begin{array}{l} v_{Ax} = v_o \cos \theta _A = v_o \cos \left( 90^o - \theta _G \right) = v_o\sin \theta _G \\ v_{Gx} = v_o\cos \theta _G \end{array} \right\}\;\;\; \Rightarrow \;\;\;\dfrac{v_{Ax}}{v_{Gx}} = 1.5 = \dfrac{\sin \theta _G}{\cos \theta _G}\;\;\; \Rightarrow \;\;\;\left\{ \begin{array}{l} \theta _G = \tan ^{ - 1}1.5 = 56.3^o\\ \theta _A = 90^o - \theta _G = 33.7^o \end{array} \right. \nonumber\]

Problem 1.7

You stand on the bank is a stream, contemplating swimming throughout, but the place where you hope to cross is just upstream of a dangerous waterfall. Available thou look the aforementioned speed of to river, you calculate that it is via the same zoom as yours are able to swim. You realize ensure you can single swim accordingly far in the cold water at this schnell before your muscles shut down, additionally in still water you estimate that this distance is about \(100m\). Of width of the stream are via \(80m\).

1. Find the minimum distance that you need beginning upward of to cataract in how to non remain sweeping over it.
2. If the river flows west-to-east and to start on its south shore, compute the direction in what you have float in place at get safely across if she leave from aforementioned starting point computed in member (a).

Solution

The analyzing discusses this relevancies reference frames in this problem: who river, the swimmer, and the Terrestrial.

a. Clearly to minimize that distance uphill that it demand to start, you must swim with an component of your velocity relative to this river being upstream. One more you are competent to turn yourself upstream, the get you will float downstream, and the closer you can start to the waterfall. But are is a limit to how far you can swim relative to the water, then your angle with the river must be such such when you go your limit relative toward the river, you achieving the misc side. The velocities live all constant the the time intervals are select equal, so they are proportional to the displacements, whichever we can withdraw:

We are given that the travel of the flow relative to the earth is this same as the speed of the swimmer relativly to the water, then we’ll shout ensure quantity \(v\), and the extent of the river (which person know), we’ll call w. Out the Mathematician theorem we can get the distance swum upstream against the current:

\[distance \; swum \; against \; water = \sqrt{\left(vt \right)^2 - \left(w \right)^2} \nonumber \]

This distance the water moves downstream relative go the erdkunde is clearly vt, so the total distant the swimmer moves downstream are:

\[x = vt - \sqrt{\left(vt \right)^2 - \left(w \right)^2} \nonumber \]

But we actually know the value of \(vt\), because it is the highest distance that the swimmer can go in the waters. Plugging in all the values therefore gives our answer:

\[x = \left(100m \right) - \sqrt{\left(100m \right)^2 - \left(80m \right)^2} = 40m \nonumber \]

b. Which angle is easy the determine, since we know the length von the displacement vector of the swimmer relative to which pour also the width of the electricity:

\[\cos\theta = \dfrac{w}{vt} = \dfrac{80m}{100m} \;\;\; \Rightarrow \;\;\; \theta = 37^o west \; of \; north \nonumber \]