Starship Enterprise and Klingon Ship Problem on Physics| University Physics

Solution:

Since the collision is barely avoided, we shall make their positions and their velocities equal. So, their velocities must be equal to the Klingon ship. For the Starship Enterprise, the final velocity will be equal to 21 km/s

$\displaystyle a=\frac{v_f-v_i}{\Delta t}$

$\displaystyle a=\frac{21-51}{t}$

$\displaystyle t=-\frac{30}{a}$

Equate their positions as well

$\displaystyle x_{enterprise}=x_{ship}$

$\displaystyle 51t+\frac{1}{2}at^2=150+21t$

Substitute $\displaystyle t=-\frac{30}{a}$ to all t’s

$\displaystyle 150-30\left(-\frac{30}{a}\right)=\frac{1}{2}a\left(-\frac{30}{a}\right)^2$

$\displaystyle 150+\frac{900}{a}=\frac{450}{a}$

$\displaystyle \frac{450}{a}=-150$

$\displaystyle a=-\frac{450}{150}$

$\displaystyle a=-3.0\:km/s^2$

Therefore, the magnitude of the acceleration should be $\displaystyle a=3.0\:km/s^2$

Calculating the unit and value of k given a velocity function| University Physics

PART A. Determine the units of k in terms of m and s.

ANSWER: $m/s^3$

We know that the unit for velocity is m/s and the unit for time is s. Therefore,

$v=kt^2$

$m/s=k\left(s\right)^2$

$k=\frac{m/s}{s^2}$

$k=m/s^3$

PART B. Determine the value of the constant k.

ANSWER: $k=49.8\:m/s^3$

$v_x=\frac{dx}{dt}$

$dx=v_xdt$

$\int \:dx=\int \:v_xdt$

$x=\int \:kt^2dt$

$x=\frac{kt^3}{3}+C$

Solve for C using the pairs $x=-7.90\:m,\:\:t=0\:s$

$C=-7.9\:$

The position function therefore is

$x\left(t\right)=\frac{kt^3}{3}-7.90$

Solve for k using the pair $x=8.70\:m,\:t=1.00\:s$

$8.70=\frac{k\left(1\right)^3}{3}-7.90$

$\frac{k}{3}=8.70+7.90$

$k=3\left(16.6\right)$

$k=49.8\:m/s^3$

Train wheels stick Problem| University Physics

What is the magnitude of the train’s acceleration after its wheel begins to stick? Assume acceleration is constant after wheel begins to stick.

ANSWER: $a=1.0\:m/s^2$

The toy train has a constant speed of $v=2\:m/s$ from $x=2\:m\:\:to\:x=6\:m$.  Then, it began to decelerate at $x=6\:m$, and finally stop at $x=8\:m$. Considering the moment after the toy train’s wheel begin to stick up to the moment when it stopped.

$\left(v_f\right)^2=\left(v_i\right)^2+2a\left(\Delta x\right)$

$0^2=\left(2\:m/s\right)^2+2a\left(8\:m-6\:m\right)$

$4a=-4$

$a=-1\:m/s^2$

The negative sign indicates that the acceleration is moving opposing the motion (a deceleration). Basically, the negative sign is just an indication of the direction of the acceleration. The magnitude of the acceleration is simple $a=1.0\:m/s^2$.

One-Dimensional Kinematics with Constant Acceleration| University Physics

Learning Goal:

To understand the meaning of the variables that appear in the equations for one-dimensional kinematics with constant acceleration.

Motion with a constant, nonzero acceleration is not uncommon in the world around us. Falling (or thrown) objects and cars starting and stopping approximate this type of motion. It is also the type of motion most frequently involved in introductory kinematics problems.

The kinematic equations for such motion can be written as

$x\left(t\right)=x_i+v_it+\frac{1}{2}at^2$

$v\left(t\right)=v_i+at,$

where the symbols are defined as follows:

• $x\left(t\right)$ is the position of the particle;
• $x_i$ is the initial position of the particle;
• $v\left(t\right)$ is the velocity of the particle;
• $v_i$ is the initial velocity of the particle;
• $a$ is the acceleration of the particle.

In answering the following questions, assume that the acceleration is constant and nonzero: a≠0.

PART B. The quantity represented by $x_i$$x_i$ is a function of time (i.e., is not constant).

Recall that $x_i$ represents an initial value, not a variable. It refers to the position of an object at some initial moment.

PART D. The quantity represented by v is a function of time (i.e., is not constant).

The velocity v always varies with time when the linear acceleration is nonzero.

PART E. Which of the given equations is not an explicit function of t and is therefore useful when you don’t know or don’t need the time?

ANSWER: $v^2=\left(v_i\right)^2+2a\left(x-x_i\right)$

PART F. A particle moves with constant acceleration a. The expression vi+at represents the particle’s velocity at what instant in time?

ANSWER: when the time t has passed since the particle’s velocity was $v_i$

More generally, the equations of motion can be written as

$x\left(t\right)=x_i+v_i\Delta t+\frac{1}{2}a\left(\Delta t\right)^2$

and

$v\left(t\right)=v_i+a\Delta t$

Here Δt is the time that has elapsed since the beginning of the particle’s motion, that is, $\Delta t=t-t_i$, where $t$ is the current time and $t_i$ is the time at which we start measuring the particle’s motion. The terms $x_i$ and $v_i$ are, respectively, the position and velocity at $t=t_i$. As you can now see, the equations given at the beginning of this problem correspond to the case $t_i=0$, which is a convenient choice if there is only one particle of interest.

What is the equation describing the position of particle B?

ANSWER: $x_B\left(t\right)=x_i+0.5v_i\left(t-t_1\right)+a\left(t-t_1\right)^2$

The general equation for the distance traveled by particle B is $x_B\left(t\right)=x_{iB}+v_{iB}\Delta t+\frac{1}{2}a_B\left(\Delta t\right)^2$ or $x_B\left(t\right)=x_{iB}+v_{iB}\left(t-t_1\right)+\frac{1}{2}a_B\left(t-t_1\right)^2$, since $\Delta t=t-t_1$ is a good choice for B. From the information given, deduce the correct values of the constants that go into the equation for $x_B\left(t\right)$ given here, in terms of A’s constants of motion.

$x_B\left(t\right)=x_i+\frac{1}{2}v_i\left(t-t_1\right)+\frac{1}{2}\left(2a\right)\left(t-t_1\right)^2$

$x_B\left(t\right)=x_i+0.5v_i\left(t-t_1\right)+a\left(t-t_1\right)^2$

PART H. At what time does the velocity of particle B equal that of particle A?

ANSWER: $t=2t_1+\frac{v_i}{2a}$

Particle A’s velocity as a function of time is $v_A\left(t\right)=v_i+at$, and particle B’s velocity as a function of time is $v_B\left(t\right)=0.5v_i+2a\left(t-t_1\right)$.

Once you have expressions for the velocities of A and B as functions of time, set them equal and find the time t at which this happens.

$v_i+at=\frac{1}{2}v_i+2a\left(t-t_1\right)$

$v_i+at=\frac{1}{2}v_i+2at-2at_1$

$2at-at=2at_1+\frac{1}{2}v_i$

$at=2at_1+\frac{1}{2}v_i$

$t=2t_1+\frac{v_i}{2a}$

What Velocity vs. Time Graphs Can Tell You| University Physics

PART A. What is the initial velocity of the particle, $v_0$$v_0$?

ANSWER: $v_0=0.5\:m/s$

The initial velocity is the velocity at t=0s. Recall that in a graph of velocity versus time, time is plotted on the horizontal axis and velocity on the vertical axis.

PART B. What is the total distance Δx traveled by the particle?

ANSWER: $\Delta x=75\:m$

Recall that the area of the region that extends over a time interval Δt under the v vs. t curve is always equal to the distance traveled in Δt. Thus, to calculate the total distance, you need to find the area of the entire region under the v vs. t curve. In the case at hand, the entire region under the v vs. t curve is not an elementary geometrical figure, but rather a combination of triangles and rectangles.

PART C. What is the average acceleration $a_{av}$$a_{av}$ of the particle over the first 20.0 seconds?

ANSWER: $a_{av}=0.075\:m/s^2$

The average acceleration of a particle between two instants of time is the slope of the line connecting the two corresponding points in a v vs. t graph.

PART D. What is the instantaneous acceleration $a$$a$ of the particle at $t=45.0\:s$$t=45.0\:s$ seconds?

ANSWER: $a=-0.20\:m/s^2$

The instantaneous acceleration of a particle at any point on a v vs. t graph is the slope of the line tangent to the curve at that point. Since in the last 10 seconds of motion, between t=40.0s and t=50.0s, the curve is a straight line, the tangent line is the curve itself. Physically, this means that the instantaneous acceleration of the particle is constant over that time interval. This is true for any motion where velocity increases linearly with time.

PART E. Which of the graphs shown below is the correct acceleration vs. time plot for the motion described in the previous parts?

Recall that whenever velocity increases linearly with time, acceleration is constant. In the example here, the particle’s velocity increases linearly with time in the first 20.0 s of motion. In the second 20.0 s , the particle’s velocity is constant, and then it decreases linearly with time in the last 10 s. This means that the particle’s acceleration is constant over each time interval, but its value is different in each interval.

In conclusion, graphs of velocity as a function of time are a useful representation of straight-line motion. If read correctly, they can provide you with all the information you need to study the motion.

Solution:

Part A

The initial velocity of the pot is zero. Find the velocity $\displaystyle v_b$ of the pot at the bottom of the window. Then using the kinematic equation that relates initial and final velocities, acceleration, and distance traveled, you can solve for the distance $\displaystyle h$.

The average velocity of the flower pot as it passes by the window is $\displaystyle v_{avg}=\frac{L_w}{t}$.

As the pot falls past your window, there will be some instant when the pot’s velocity equals the average velocity $\displaystyle v_{avg}$. Recall that, under constant acceleration, velocity changes linearly with time. This means that the average velocity during a time interval will occur at the middle of that time interval. Meaning, the average velocity happened at time, $\displaystyle \frac{t}{2}$.

Considering the motion at the middle and at the bottom of the window.

$\displaystyle g=\frac{v_b-v_{avg}}{\frac{t}{2}}$

$\displaystyle v_b=\frac{gt}{2}+v_{ave}$

$\displaystyle v_b=\frac{gt}{2}+\frac{L_w}{t}$

So, we now know that velocity at the bottom of the window. Consider the motion from the top (point of dropped) to the bottom of the window

$\displaystyle \left(v_b\right)^2-\left(v_0\right)^2=2g\left(h-0\right)$

$\displaystyle \left(v_b\right)^2=2gh$

$\displaystyle h=\frac{\left(v_b\right)^2}{2g}$

$\displaystyle h=\frac{\left(\frac{gt}{2}+\frac{L_w}{t}\right)^2}{2g}=\frac{\left(\frac{gt^2+2L_w}{2t}\right)^2}{2g}=\frac{\left(gt^2+2L_w\right)^2}{8gt^2}$

Part B

$\displaystyle \left(v_{ground}\right)^2-\left(v_b\right)^2=2gh_b$

$\displaystyle \left(v_{ground}\right)^2=2gh_b+\left(v_b\right)^2$

$\displaystyle\left(v_{ground}\right)=\sqrt{2gh_b+\left(v_b\right)^2}$

To demonstrate the tremendous acceleration of a top fuel Drag Racer| University Physics

PART A. What is $\displaystyle t_{max}$$\displaystyle t_{max}$, the longest time after the dragster begins to accelerate that you can possibly run into the back of the dragster if you continue at your initial velocity?

$\displaystyle t_{max}=\frac{v_0}{a}$

PART B. Assuming that the dragster has started at the last instant possible (so your front bumper almost hits the rear of the dragster at $\displaystyle t=t_{max}$$\displaystyle t=t_{max}$), find your distance from the dragster when he started. If you calculate positions on the way to this solution, choose coordinates so that the position of the drag car is 0 at t=0. Remember that you are solving for a distance (which is a magnitude, and can never be negative), not a position (which can be negative).

$\displaystyle D_{start}=\frac{\left(v_0\right)^2}{2a}$

$\displaystyle D_{start}=D_{car}-x_d\left(t_{max}\right)$

$\displaystyle D_{start}=v_0t_{max}-\frac{1}{2}a\left(t_{max}\right)^2$

$\displaystyle D_{start}=v_0\left(\frac{v_0}{a}\right)-\frac{1}{2}a\left(\frac{v_0}{a}\right)^2$

$\displaystyle D_{start}=\frac{\left(v_0\right)^2}{a}-\frac{\left(v_0\right)^2}{2a}$

$\displaystyle D_{start}=\frac{\left(v_0\right)^2}{2a}$

PART C. Find numerical values for $\displaystyle t_{max}$$\displaystyle t_{max}$ and $\displaystyle D_{start}$$\displaystyle D_{start}$ in seconds and meters for the (reasonable) values $\displaystyle v_0=60\:mph$$\displaystyle v_0=60\:mph$ (26.8 m/s) and $\displaystyle a=50\:m/s^2$$\displaystyle a=50\:m/s^2$

$\displaystyle t_{max}=\frac{v_0}{a}=\frac{26.8\:m/s}{50\:m/s^2}=0.54\:s$

$\displaystyle D_{start}=\frac{\left(v_0\right)^2}{2a}=\frac{\left(26.8\:m/s\right)^2}{2\left(50\:m/s^2\right)}=7.2\:m$

The blue curve shows how the car, initially at $\displaystyle x_0$, continues at constant velocity (blue) and just barely touches the accelerating drag car (red) at $\displaystyle t_{max}$.

Running and Walking Problem| University Physics

PART A. How long does it take Rick to cover the distance D?

Find the time that it takes Rick to walk the first half of the distance, that is, to travel a distance D/2 at speed $\displaystyle v_w$.

$\displaystyle t_{w,R}=\frac{D}{2v_w}$

Now find the time Rick spends running.

$\displaystyle t_{r,R}=\frac{D}{2v_r}$

Now just add the two times up and you’re done.

$\displaystyle t_R=\frac{D}{2v_w}+\frac{D}{2v_r}=\frac{D}{2v_wv_r}\left(v_w+v_r\right)$

PART B. Find Rick’s average speed for covering the distance D.

You were given the total distance and have calculated the total time. Recall that average speed is equal to the total distance traveled divided by the amount of time it took to travel this distance.

$\displaystyle v_{ave,\:R}=\frac{2v_rv_w}{v_w+v_r}$

PART C. How long does it take Tim to cover the distance?

Tim walks at speed $\displaystyle v_w$ half the time and runs at speed $\displaystyle v_r$ for the other half.

$\displaystyle v_{ave,\:T}=\frac{v_w+v_r}{2}$

The time is just the distance divided by the average speed.

$\displaystyle t_T=\frac{D}{\frac{v_w+v_r}{2}}=\frac{2D}{v_r+v_w}$

PART D. Who covers the distance D more quickly?

Imagine that both Rick and Tim do all of their walking before they start to run. Rick will start running when he has covered half of the total distance. When Tim reaches half of the total distance, will he already have started running?

PART E. In terms of given quantities, by what amount of time, Δt, does Tim beat Rick?

$\displaystyle \Delta t=\frac{D\left(v_w-v_r\right)^2}{2v_rv_w\left(v_r-v_w\right)}$

This is just simple subtraction between the two computed times.

PART F. In the special case that vr=vw, what would be Tim’s margin of victory Δt(vr=vw)?

If vr=vw, is the any difference between what Tim and Rick do?

Half the Distance and Half the Time Problem| University Physics

PART A. What is Julie’s average speed on the way to Grandmother’s house?

Julie drove 50 miles at a speed of 35 mph, and drove another 50 miles for 65 mph. So, for the first 50 miles, she drove for

$\displaystyle time=\frac{distance}{speed}=\frac{50\:mi}{35\:mph}=\frac{10}{7}\:hours$

and for the next 50 miles, she drove for

$\displaystyle time=\frac{distance}{speed}=\frac{50\:mi}{65\:mph}=\frac{10}{13}\:hours$

Therefore, her average speed was

$\displaystyle average\:speed=\frac{total\:distance}{total\:time}$

$\displaystyle average\:speed=\frac{100\:miles}{\frac{10}{7}+\frac{10}{13}\:hours}=45.5\:mph$

PART B. What is her average speed on the return trip?

Since the time she used driving at 35 mph is the same amount of time she used driving at 65 mph, the average speed is just the average of the two speeds given.

Position, Velocity, and Acceleration values from Position Function| University Physics

PART A.

Evaluate the position at time t= 3.00 s.

$\displaystyle x=2.00\left(3\right)^3-5.00\left(3\right)+3=42.0\:m$

PART B.

Determine the velocity function v(t) from the position function x(t) by differentiation.

$\displaystyle v\left(t\right)=6t^2-5$

$\displaystyle v\left(t\right)=6\left(3\right)^2-5=49.0\:m/s$

PART C.

$\displaystyle a\left(t\right)=12t$
$\displaystyle a\left(t\right)=12t=12\left(3\right)=36\:m/s^2$