College Physics 2.66 – Solving for velocity and acceleration from position graph


Figure 2.68 shows the position graph for a particle for 6 s.

(a) Draw the corresponding Velocity vs. Time graph.

(b) What is the acceleration between 0 s and 2 s?

(c) What happens to the acceleration at exactly 2 s?

position graph for a particle for 6 s.
Figure 2.68

Solution:

Part A

The velocity of the particle is the slope of the position vs time graph. Since the position graph is compose of straight lines, we can say that the velocity is constant for several time ranges.

Time RangesSlope of the position Graph/Velocity
0 to 2 seconds=\frac{2-0}{2-0}=1\:\text{m/s}
2 to 3 seconds=\frac{-3-2}{3-2}=\frac{-5}{1}=-5\:\text{m/s}
3 to 5 seconds0 \text{m/s}
5 to 5 seconds=\frac{-2-\left(-3\right)}{6-5}=\frac{1}{1}=1\:\text{m/s}

Based from the data in the table, we can draw the velocity diagram

velocity vs time graph

Part B

Since the velocity is constant between 0 seconds and 2 seconds, we say that the acceleration is 0.

Part C

Since there is a sudden change in velocity at exactly 2 seconds in a very short amount of time, we say that the acceleration is undefined in this case.


Starship Enterprise and Klingon Ship Problem on Physics| University Physics


The Starship Enterprise returns from warp drive to ordinary space with a forward speed of 51 km/s. To the crew’s great surprise, a Klingon ship is 150 km directly ahead, traveling in the same direction at a mere 21 km/s. Without evasive action, the Enterprise will overtake and collide with the Klingons in just about 5.0 s. The Enterprise’s computers react instantly to brake the ship.

PART A. What magnitude acceleration does the Enterprise need to just barely avoid a collision with the Klingon ship? Assume the acceleration is constant.


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


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

A particle’s velocity is described by the function v_x=kt^2 where v_x is in m/s, t is in s, and k is a constant. The particle’s position at t_0=0 is x_0=-7.90\:m. At t_1=1.00\:s, the particle is at x_1=8.70\:m.

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

A toy train is pushed forward and released at x_0=2.0\:m with a speed of 2.0 m/s. It rolls at a steady speed for 2.0 s, then one wheel begins to stick. The train comes to a stop 6.0 m from the point at which it was released.

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 A. The quantity represented by x is a function of time (i.e., is not constant).

ANSWER: True

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

ANSWER: False

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

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

ANSWER: False

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

ANSWER: True

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.

PART G. To illustrate the use of these more general equations, consider the motion of two particles, A and B. The position of particle A depends on time as x_A\left(t\right)=x_i+v_it+\frac{1}{2}at^2. That is, particle A starts moving at time t=t_{iA}=0 with velocity v_{iA}=v_i, from x_{iA}=x_i. At time t=t_1, particle B has twice the acceleration, half the velocity, and the same position that particle A had at time t=0.

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

A common graphical representation of motion along a straight line is the v vs. t graph, that is, the graph of (instantaneous) velocity as a function of time. In this graph, time t is plotted on the horizontal axis and velocity v on the vertical axis. Note that by definition, velocity and acceleration are vector quantities. In a straight-line motion, however, these vectors have only a single nonzero component in the direction of motion. Thus, in this problem, we will call v the velocity and a the acceleration, even though they are really the components of the velocity and acceleration vectors in the direction of motion, respectively.

Here is a plot of velocity versus time for a particle that travels along a straight line with varying velocity. Refer to this plot to answer the following questions.

The plot shows the x component of the velocity as a function of time. Time is plotted from 0 to 50 seconds on the x-axis. The x component of the velocity is plotted from 0 to 2 meters per second on the y-axis. The velocity starts at 0 seconds and 0.5 meters per second, then increases linearly to 2.0 meters per second at 20 seconds and stays at this value until 40 seconds. Then it decreases linearly to 0 at 50 seconds.

PART A. What is the initial velocity of the particle, 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} 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 of the particle at 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?

There are four graphs showing the acceleration as a function of time. Time ranges from 0 to 50 seconds on the horizontal axis. The acceleration is from -0.20 to 0.2 meters per second squared on the vertical axis. On graph A, the acceleration equals 0.075 meters per second squared through the first 40 seconds, then equals -0.20 meters per second squared through the next 10 seconds. On graph B, the acceleration equals 0.075 meters per second squared through the first 20 seconds, then equals zero through the next 20 seconds and then equals -0.20 meters per second squared through the next 10 seconds. On graph C, the acceleration equals 0.2 meters per second squared through the first 20 seconds, then equals zero through the next 20 seconds. Then, it decreases linearly to -0.20 meters per second squared through the next 10 seconds. On graph D, the acceleration equals -0.075 meters per second squared through the first 20 seconds, then equals zero through the next 20 seconds and then equals to 0.075 meters per second squared through the next 10 seconds.

ANSWER: Graph B

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.

A Flower Pot Falling Past a Window| University Physics


As you look out of your dorm window, a flower pot suddenly falls past. The pot is visible for a time \displaystyle  t, and the vertical length of your window is \displaystyle L_w. Take down to be the positive direction, so that downward velocities are positive and the acceleration due to gravity is the positive quantity \displaystyle g.

Assume that the flower pot was dropped by someone on the floor above you (rather than thrown downward).

PART A. From what height \displaystyle h above the bottom of your window was the flower pot dropped?

PART B. If the bottom of your window is a height \displaystyle h_b above the ground, what is the velocity \displaystyle v_{ground} of the pot as it hits the ground? You may introduce the new variable \displaystyle v_b, the speed at the bottom of the window, defined by

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


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}


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


To demonstrate the tremendous acceleration of a top fuel drag racer, you attempt to run your car into the back of a dragster that is “burning out” at the red light before the start of a race. (Burning out means spinning the tires at high speed to heat the tread and make the rubber sticky.)

You drive at a constant speed of \displaystyle v_0 toward the stopped dragster, not slowing down in the face of the imminent collision. The dragster driver sees you coming but waits until the last instant to put down the hammer, accelerating from the starting line at constant acceleration, \displaystyle a. Let the time at which the dragster starts to accelerate be \displaystyle t=0.

The figure shows a car moving to the right with speed v 0 toward a stopped dragster, which stands in front of the lights facing to the right.

PART A. What is \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?


Answer:

\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}), 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).


Answer:

\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} and \displaystyle D_{start} in seconds and meters for the (reasonable) values \displaystyle v_0=60\:mph (26.8 m/s) and \displaystyle a=50\:m/s^2


Answer:

\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}.


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Running and Walking Problem| University Physics


Tim and Rick both can run at speed \displaystyle v_r and walk at speed \displaystyle v_w , with \displaystyle v_r>v_w. They set off together on a journey of distance D. Rick walks half of the distance and runs the other half. Tim walks half of the time and runs the other half.

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


Answer:

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.


Answer:

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?


Answer:

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?


Answer: Tim

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?


Answer:

\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)?


ANSWER: 0

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


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Half the Distance and Half the Time Problem| University Physics


Julie drives 100 mi to Grandmother’s house. On the way to Grandmother’s, Julie drives half the distance at 35.0 mph and half the distance at 65.0 mph. On her return trip, she drives half the time at 35.0 mph and half the time at 65.0 mph.

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


ANSWER: 45.5 mph

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?


ANSWER: 50.0 m/s

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.


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