College Physics 2.64 – Consistency on the position and velocity diagrams


(a) Take the slope of the curve in Figure 2.64 to find the jogger’s velocity at t = 2.5 s.

(b) Repeat at 7.5 s. These values must be consistent with the graph in Figure 2.65.

Figure 2.64
Figure 2.65

Solution:

The velocity at time 2.5 seconds is 5 m/s.

The velocity at time 7.5 seconds is -4 m/s.


College Physics 2.63 – Constructing a displacement graph


Construct the displacement graph for the subway shuttle train as shown in Figure 2.18(a). Your graph should show the position of the train, in kilometers, from t = 0 to 20 s. You will need to use the information on acceleration and velocity given in the examples for this figure.

2.63


Solution:

The position vs time graph is shown in the figure below. 

2.63b

College Physics 2.62 – Acceleration as the slope of the velocity vs time diagram


By taking the slope of the curve in Figure 2.63, verify that the acceleration is 3.2 m/s²  at t = 10 s.

2.62
Figure 2.63

Solution:

Since the graph is a straight line, we can use the two points before and after the specified time to determine the slope of the line. The slope of the velocity-time graph is the acceleration.

The two points are \left(0,\:165\right)\:and\:\left(20,\:228\right)

The velocityis computed as

a=m=\frac{\Delta y}{\Delta x}

a=\frac{y_2-y_2}{x_2-x_1}

a=\frac{228\:m/s-165\:m/s}{20\:s-0\:s}

a=3.2\:m/s^2

Therefore, the acceleration is verified to be 3.2 m/s² at t=10 s.


College Physics 2.61 – Velocity as slope of position vs time


Using approximate values, calculate the slope of the curve in Figure 2.62 to verify that the velocity at t = 30.0 s is 0.238 m/s. Assume all values are known to 3 significant figures.

2.60
Figure 2.62

Solution:

Since the graph is a straight line, we can use any two points to determine the slope of the line. The slope of the position-time graph is the velocity.

The two points are \left(20,\:6.95\right)\:and\:\left(40,\:11.7\right)

The velocityis computed as

v=m=\frac{\Delta y}{\Delta x}

v=\frac{y_2-y_2}{x_2-x_1}

v=\frac{11.7\:km-6.95\:km}{40\:s-20\:s}

v=0.238\:km/s

Therefore, the velocity is 0.238 km/s.


College Physics 2.60 – Slope of the position vs time diagram


Using approximate values, calculate the slope of the curve in Figure 2.62 to verify that the velocity at t = 10.0 s is 0.208 km/s. Assume all values are known to 3 significant figures.

2.60
Figure 2.62

Solution:

Since the graph is a straight line, we can use any two points to determine the slope of the line. The slope of the position-time graph is the velocity.

The two points are \left(0,\:2.80\right)\:and\:\left(20,\:6.95\right)

The velocityis computed as

v=m=\frac{\Delta y}{\Delta x}

v=\frac{y_2-y_2}{x_2-x_1}

v=\frac{6.95\:km-2.80\:km}{20\:s-0\:s}

v=0.208\:km/s

Therefore, the velocity is 0.208 km/s.


College Physics 2.59 – Analysis of motion diagrams


(a) By taking the slope of the curve in Figure 2.60, verify that the velocity of the jet car is 115 m/s at t = 20 s.

(b) By taking the slope of the curve at any point in Figure 2.61, verify that the jet car’s acceleration is 5.0 m/s².

Figure 2.60
Figure 2.61

Solution:

Part A

Based from the graph shown in figure 2.60, we can choose two points, one before and one after t=20 s. We have the points \left(15,\:988\right)\:and\:\left(25,\:2138\right).

The slope is computed using the formula:

m=\frac{\Delta y}{\Delta x}

m=\frac{2138\:m-988\:m}{25\:s-15\:s}

m=115\:m/s

Therefore, we have verified that the velocity of the jet car is 115 m/s at t=20 seconds. 

Part B

Since the graph is a straight line, we can use any two points to determine the slope of the line. The slope of the velocity-time graph is the acceleration.

The two points are \left(10,\:65\right)\:and\:\left(25,\:140\right)

The acceleration is computed as

a=m=\frac{\Delta y}{\Delta x}

a=\frac{y_2-y_2}{x_2-x_1}

a=\frac{140\:m/s-65\:m/s}{25\:s-10\:s}

a=5\:m/s^2

Therefore, the acceleration is verified to be 5.0 m/s².


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}