Tag Archives: motion in one dimension

Physics for Scientists and Engineers 3E by R. Knight, P1.22


Draw a pictorial representation for the following problem. Do not solve the problem. What acceleration does a rocket need to reach a speed of 200 m/s at a height of 1.0 km?


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Physics for Scientists and Engineers 3E by R. Knight, C1E&P21


Draw a pictorial representation for the following problem. Do not solve the problem. The light turns green, and a bicyclist starts forward with an acceleration of 1.5 m/s2. How far must she travel to reach a speed of 7.5 m/s.


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Physics for Scientists and Engineers 3E by R. Knight, C1E&P20


Write a short description of the motion of a real object for which Figure EX1.20 would be a realistic position-versus-time graph.

Figure 1.20 Physics for Scientists and Engineers Third Edition by Randall Knight
Figure EX1.20

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Physics for Scientists and Engineers 3E by R. Knight, C1E&P19


Write a short description of the motion of a real object for which Figure EX1.19 would be a realistic position-versus-time graph.

Figure 1.19 Physics for Scientists and Engineers Third Edition by Randall Knight
Figure EX1.19

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Physics for Scientists and Engineers 3E by R. Knight, C1E&P18


Figure EX1.18 shows the motion diagram of a drag racer. The camera took one frame every 2 s.

Figure 1.18 Physics for Scientists and Engineers Third Edition by Randall Knight
Figure EX1.18

  1. Measure the x-value of the racer at each dot. List your data in a table similar to Table 1.1, showing each position and the time at which it occurred.
  2. Make a position-versus-time graph for the drag racer. Because you have data only at certain instants, your graph should consist of dots that are not connected together.


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Grantham PHY220 Week 2 Assignment Problem 8

If a car is traveling at 50 m/s and then stops over 300 meters (while sliding), what is the coefficient of kinetic friction between the tires of the car and the road?

SOLUTION:

Draw the free-body diagram of the car

week 2 problem 8

Consider the vertical direction

\sum F_y=ma_y

F_N-mg=0

F_N=mg

Consider the motion in the horizontal direction

Solve for the acceleration of the car.

v^2=\left(v_0\right)^2+2a_x\Delta x

a_x=\frac{v^2-\left(v_0\right)^2}{2\Delta x}=\frac{0-50^2}{2\left(300\right)}=-4.17\:m/s^2

Solve for the coefficient of kinetic friction

\sum F_x=ma_x

-F_{fr}=ma_x

-\mu _kF_N=ma_x

\mu _k=\frac{ma_x}{-F_N}=\frac{m\:\left(-4.17\right)}{-m\left(9.80\right)}=\frac{4.17}{9.80}

\mu _k=0.43

College Physics by Openstax Chapter 2 Problem 30


A fireworks shell is accelerated from rest to a velocity of 65.0 m/s over a distance of 0.250 m.

(a) How long did the acceleration last?

(b) Calculate the acceleration.


Solution:

We are given the following values:v_0=0\:\text{m/s}; v_f=65.0\:\text{m/s}; and \Delta x=0.250\:\text{m}.

We can immediately solve for the acceleration using the given values, so we are going to answer Part B first.

Part B

Solve for the acceleration first using the formula

\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

We solve for acceleration in terms of the other variables.

a=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}

Substitute the given values

\begin{align*}
a & =\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x} \\
a & = \frac{\left(65.0\:\text{m/s}\right)^2-\left(0\:\text{m/s}\right)^2}{2\left(0.250\:\text{m}\right)} \\
a & =8450\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part A

To solve for the time of this motion, we shall use the formula

v_f=v_0+at

Solving for time, t, in terms of the other variables we have.

t=\frac{v_f-v_0}{a}

We now substitute the values given, and the computed acceleration to find the time.

\begin{align*}
t & =\frac{v_f-v_0}{a} \\
t & =\frac{65.0\:\text{m/s}-0\:\text{m/s}}{8450\:\text{m/s}^2} \\
t & =7.6922\:\times 10^{-3}\:\text{s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

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College Physics by Openstax Chapter 2 Problem 28


A powerful motorcycle can accelerate from rest to 26.8 m/s (100 km/h) in only 3.90 s.

(a) What is its average acceleration?

(b) How far does it travel in that time?


Solution:

We are given the following: v_0=0 \ \text{m/s}; v_f=26.8 \ \text{m/s}; and t=3.90\ \text{s}.

Part A

The average acceleration of the motorcycle can be solved using the equation \overline{a}=\frac{\Delta v}{\Delta t}. Substitute the given into the equation. That is,

\begin{align*}
\overline{a} & =\frac{\Delta v}{\Delta t} \\
\overline{a} & =\frac{26.8\:\text{m/s}-0\:\text{m/s}}{3.90\:\text{s}} \\
\overline{a} & =6.872\:\text{m/s}^2\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

The distance traveled is equal to the average velocity multiplied by the time of travel. That is,

\begin{align*}
\Delta x & =v_{ave}t\\
\Delta x & =\left(\frac{0\:\text{m/s}+26.8\:\text{m/s}}{2}\right)\left(3.90\:\text{s}\right) \\
\Delta x & =52.26\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

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College Physics by Openstax Chapter 2 Problem 27


In a slap shot, a hockey player accelerates the puck from a velocity of 8.00 m/s to 40.0 m/s in the same direction. If this shot takes 3.33×10-2 s, calculate the distance over which the puck accelerates.


Solution:

The best equation that can be used to solve this problem is \Delta x=v_{ave} t. That is,

\begin{align*}
\Delta x & = v_{ave} t \\
\Delta x & = \left(\frac{8\:\text{m/s}+40\:\text{m/s}}{2}\right)\left(3.33\times 10^{-2}\:\text{s}\right) \\
\Delta x & = 0.7992\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, the distance over which the puck accelerates is 0.7992 meters.


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College Physics by Openstax Chapter 2 Problem 26


Blood is accelerated from rest to 30.0 cm/s in a distance of 1.80 cm by the left ventricle of the heart.

(a) Make a sketch of the solution.

(b) List the knowns in this problem.

(c) How long does the acceleration take? To solve this part, identify the unknown, and then discuss how you chose the appropriate equation to solve for it. After choosing the equation, show your steps in solving for the unknown, checking your units.

(d) Is the answer reasonable when compared with the time for a heartbeat?


Solution:

Part A

The sketch should contain the starting point and the final point. This will be done by connecting a straight line from the starting point to the final point. The sketch is shown below.

Part B

The list of known variables are:

Initial velocity: v_0=0\:\text{m/s}
Final Velocity: v_f=30.0\:\text{cm/s}
Distance Traveled: x-x_0=1.80\:\text{cm}

Part C

The best equation to solve for this is \Delta \text{x}=\text{v}_{\text{ave}}\text{t} where v_{ave} is the average velocity, and t is time. That is

\begin{align*}
\Delta x & =v_{ave} t \\
t &=\frac{\Delta x}{v_{ave}} \\
t & =\frac{1.80\:\text{cm}}{\frac{\left(0\:\text{cm/s}+30\:\text{cm/s}\right)}{2}}\\
t & =0.12\:\text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

Part D

Since the computed value of the time for the acceleration of blood out of the ventricle is only 0.12 seconds (only a fraction of a second), the answer seems reasonable. This is due to the fact that an entire heartbeat cycle takes about one second. So, the answer is yes, the answer is reasonable.


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