## Solution:

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

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

## Solution:

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

## 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.

## 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.

## 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.

## 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

### 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.

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

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

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

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

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