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=−2.10×104 m/s2;t=1.85×10−3 s;vf=0 m/sa=-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

v0=vf−atv_0=v_f-at

Substitute the given values

v0=0 m/s−(−2.10×104 m/s2)(1.85×10−3 s)v0=38.85 m/s  (Answer)\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 a‾=4.50 m/s2, Δt=2.40 sec⁡, and v0=0 m/s\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 vfv_f. The formula in solving for vfv_f is

vf=v0+atv_f=v_0+at

Substituting the given values,

vf=0 m/s+(4.50 m/s2)(2.40 s)vf=108 m/s  (Answer)\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=v0t+12at2x=v_0t+\frac{1}{2}at^2

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

x=0+12(4.50 m/s2)(t2)x=2.52t2\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 

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

Substituting the given values

a‾=vf−v0Δta‾=6.5×103 m/s−0 m/s60.0 seca‾=108.33 m/s2  (Answer)\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

a‾=108.33 m/s29.80 m/s2a‾=11.05g  (Answer)\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

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

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

Δt=Δva‾\Delta t=\frac{\Delta v}{\overline{a}}

Substituting the given values, we have

Δt=Δva‾Δt=vf−v0a‾Δt=2.00 m/s−0 m/s1.40 m/s2Δt=1.43 seconds  (Answer)\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

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

Then substituting all the given values, we have

a‾=vf−v0Δta‾=0 m/s−2 m/s0.8 m/s2a‾=−2.50 m/s2  (Answer)\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

a‾=ΔvΔta‾=vf−v0tf−t0\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

a‾=282 m/s−0 m/s5.00 sec⁡a‾=56.4 m/s2  (Answer)\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 

a‾=0 m/s−282 m/s1.40 sa‾=−201.43 m/s2  (Answer)\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

a‾=56.49.80=5.76g  (Answer)\overline{a}=\frac{56.4}{9.80}=5.76\text{g} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

The deceleration is

a‾=201.439.80=20.55g  (Answer)\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 

a‾=change in velocitychange in timea‾=ΔvΔta‾=vf−v0tf−t0\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

a‾=30.0 m/s−0 m/s7.00 s=4.29 m/s2  (Answer)\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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