Tag Archives: acceleration

College Physics by Openstax Chapter 2 Problem 21

A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 2.10×104 m/s2, and 1.85 ms (1 ms = 10-3 s) elapses from the time the ball first touches the mitt until it stops, what was the initial velocity of the ball?

Solution:

We are given the following values: a=-2.10 \times 10^4 \ \text{m/s}^2; t=1.85 \times 10^{-3} \ \text{s}; v_f=0 \ \text{m/s}.

The formula in solving for the initial velocity is

v_0=v_f-at

Substitute the given values

\begin{align*}
v_0 & =0\:\text{m/s}-\left(-2.10\times 10^4\text{ m/s}^2\right)\left(1.85\times 10^{-3}\:\text{s}\right) \\
v_0 & =38.85\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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College Physics by Openstax Chapter 2 Problem 20

An Olympic-class sprinter starts a race with an acceleration of 4.50 m/s2.

(a) What is her speed 2.40 s later?

(b) Sketch a graph of her position vs. time for this period.

Solution:

We are given \overline{a}=4.50\:\text{m/s}^2, \ \Delta t=2.40\:\sec ,\:\text{and}\: v_0=0\:\text{m/s}

Part A

The unknown is v_f. The formula in solving for v_f is

v_f=v_0+at

Substituting the given values,

\begin{align*}
v_f & =0\:\text{m/s}+\left(4.50\:\text{m/s}^2\right)\left(2.40\:\text{s}\right) \\
v_f & = 108\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

The relationship between position and time can be calculated using the formula

x=v_0t+\frac{1}{2}at^2

Then, with the given, we can express position in terms of time

\begin{align*}
x & =0+\frac{1}{2}\left(4.50\:\text{m/s}^2\right)\left(\text{t}^2\right) \\
x & =2.52\text{t}^2 \\
\end{align*}

The values of the position given the time are tabulated below

[wpdatatable id=2]

The values are plotted in the coordinate axes 

Time vs Position: College Physics 2.20 - Acceleration of an Olympic-class Sprinter
Time vs Position
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College Physics by Openstax Chapter 2 Problem 19

Assume that an intercontinental ballistic missile goes from rest to a suborbital speed of 6.50 km/s in 60.0 s (the actual speed and time are classified). What is its average acceleration in m/s2 and in multiples of g (9.80 m/s2) ?

Solution:

The formula for acceleration is 

\overline{a}=\frac{\Delta v}{\Delta t}

Substituting the given values

\begin{align*}
\overline{a} & = \frac{v_f-v_0}{\Delta t} \\
\overline{a} & =\frac{6.5\times 10^3\:\text{m/s}-0\:\text{m/s}}{60.0\:\text{sec}}\\
\overline{a} & =108.33\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

This can be expressed in multiples of g

\begin{align*}
\overline{a} & = \frac{108.33\:\text{m/s}^2}{9.80\:\text{m/s}^2}\\
\overline{a} &  =11.05\text{g} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Therefore, the average acceleration is 108.33 m/s2 and can be expressed as 11.05g.

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College Physics by Openstax Chapter 2 Problem 18

A commuter backs her car out of her garage with an acceleration of 1.40 m/s2 .

(a) How long does it take her to reach a speed of 2.00 m/s?

(b) If she then brakes to a stop in 0.800 s, what is her deceleration?

Solution:

Part A

The formula for acceleration is

\overline{a}=\frac{\Delta v}{\Delta t}

If we rearrange the formula by solving for \Delta t, in terms of velocity and acceleration, we come up with

\Delta t=\frac{\Delta v}{\overline{a}}

Substituting the given values, we have

\begin{align*}
\Delta t & =\frac{\Delta v}{\overline{a}} \\
\Delta t & = \frac{v_f-v_0}{\overline{a}} \\
\Delta t & =\frac{2.00 \ \text{m/s}-0 \ \text{m/s}}{1.40 \ \text{m/s}^2} \\
\Delta t & =1.43 \ \text{seconds} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

The formula for acceleration (deceleration) is

\overline{a}=\frac{\Delta v}{\Delta t}

Then substituting all the given values, we have

\begin{align*}
\overline{a} & = \frac{v_f-v_0}{\Delta t} \\
\overline{a} & = \frac{0 \ \text{m/s}-2\ \text{m/s}}{0.8 \ \text{m/s}^2} \\
\overline{a} & = -2.50 \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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College Physics by Openstax Chapter 2 Problem 17

Dr. John Paul Stapp was U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s! Calculate his

(a) acceleration and

(b) deceleration.

Express each in multiples of g (9.80 m/s2) by taking its ratio to the acceleration of gravity.

Solution:

Part A

The formula for acceleration is

\begin{align*}
\overline{a} & =\frac{\Delta v}{\Delta t} \\
\overline{a} & = \frac{v_f-v_0}{t_f-t_0} \\
\end{align*}

Substituting the given values

\begin{align*}
\overline{a} & =\frac{282\:\text{m/s}-0\:\text{m/s}}{5.00\:\sec } \\
\overline{a} & =56.4\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

The deceleration is 

\begin{align*}
\overline{a} & =\frac{0\:\text{m/s}-282\:\text{m/s}}{1.40\:\text{s}} \\
\overline{a} & =-201.43\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

In expressing the computed values in terms of g, we just divide them by 9.80.

The acceleration is

\overline{a}=\frac{56.4}{9.80}=5.76\text{g} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

The deceleration is

\overline{a}=\frac{201.43}{9.80}=20.55\text{g} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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College Physics by Openstax Chapter 2 Problem 16

A cheetah can accelerate from rest to a speed of 30.0 m/s in 7.00 s. What is its acceleration?

Solution:

The formula for acceleration is 

\begin{align*}
\overline{a} & =\frac{\text{change in velocity}}{\text{change in time}}\\
\overline{a}  & =\frac{\Delta \text{v}}{\Delta t} \\
\overline{a}  & =\frac{v_f-v_0}{t_f-t_0} \\
\end{align*}

Substituting the given values

\begin{align*}
\overline{a} & =\frac{30.0\:\text{m/s}-0\:\text{m/s}}{7.00\:\text{s}} \\
& =4.29\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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