Tag Archives: Velocity

College Physics by Openstax Chapter 2 Problem 66

Figure 2.68 shows the position graph for a particle for 6 s. (a) Draw the corresponding Velocity vs. Time graph. (b) What is the acceleration between 0 s and 2 s? (c) What happens to the acceleration at exactly 2 s?

position graph for a particle for 6 s.
Figure 2.68

Solution:

Part A

The velocity of the particle is the slope of the position vs time graph. Since the position graph is composed of straight lines, we can say that the velocity is constant for several time ranges.

Time RangeSlope of the Position vs Time Graph
0 to 2 seconds=2−02−0=1 m/s=\frac{2-0}{2-0}=1\:\text{m/s}
2 to 3 seconds=−3−23−2=−51=−5 m/s=\frac{-3-2}{3-2}=\frac{-5}{1}=-5\:\text{m/s}
3 to 5 seconds=0 m/s=0 \ \text{m/s}
5 to 6 seconds=−2−(−3)6−5=11=1 m/s=\frac{-2-\left(-3\right)}{6-5}=\frac{1}{1}=1\:\text{m/s}

Based on the data in the table, we can draw the velocity diagram

velocity vs time graph
velocity vs time graph

Part B

Since the velocity is constant between 0 seconds and 2 seconds, we say that the acceleration is 0.

Part C

Since there is a sudden change in velocity at exactly 2 seconds in a very short amount of time, we say that the acceleration is undefined in this case.

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College Physics by Openstax Chapter 2 Problem 43

A basketball referee tosses the ball straight up for the starting tip-off. At what velocity must a basketball player leave the ground to rise 1.25 m above the floor in an attempt to get the ball?

Solution:

It is our assumption that the player attempts to get the ball at the top where the velocity is zero.

The given are the following: vfy=0 m/sv_{fy}=0 \ \text{m/s}; Δy=1.25 m\Delta y=1.25 \ \text{m}; and a=−9.80 m/s2a=-9.80 \ \text{m/s}^2.

We are required to solve for the initial velocity v0yv_{0y} of the player. We are going to use the formula

(vfy)2=(voy)2+2aΔy\left(v_{fy}\right)^2=\left(v_{oy}\right)^2+2a\Delta y

Solving for voyv_{oy} in terms of the other variables:

voy=(vfy)2−2aΔyv_{oy}=\sqrt{\left(v_{fy}\right)^2-2a\Delta y}

Substituting the given values:

voy=(vfy)2−2aΔyvoy=(0 m/s)2−2(−9.80 m/s2)(1.25 m)voy=4.95 m/s  (Answer)\begin{align*} v_{oy} & =\sqrt{\left(v_{fy}\right)^2-2a\Delta y} \\ v_{oy} & = \sqrt{\left(0\:\text{m/s}\right)^2-2\left(-9.80\:\text{m/s}^2\right)\left(1.25\:\text{m}\right)} \\ v_{oy} & =4.95 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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College Physics by Openstax Chapter 2 Problem 42

Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, (d) 2.00, and (e) 2.50 s for a rock thrown straight down with an initial velocity of 14.0 m/s from the Verrazano Narrows Bridge in New York City. The roadway of this bridge is 70.0 m above the water.

Solution:

The given known quantities are: a=−9.8 m/s2a=-9.8\:\text{m/s}^2; y0=0 my_0=0 \ \text{m}; and v0=−14 m/sv_0=-14 \ \text{m/s}.

To compute for the displacement, we use the formula

Δy=v0t+12at2\Delta y=v_0t+\frac{1}{2}at^2

and to compute for the final velocity, we use the formula

vf=v0+atv_f=v_0+at

Part A

The displacement at t=0.500 st=0.500 \ \text{s} is

Δy=v0t+12at2Δy=(−14.0 m/s)(0.500 s)+12(−9.8 m/s2)(0.500 s)2Δy=−8.23 m  (Answer)\begin{align*} \Delta y & =v_0t+\frac{1}{2}at^2 \\ \Delta y &=\left(-14.0\:\text{m/s}\right)\left(0.500\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(0.500\:\text{s}\right)^2 \\ \Delta y & =-8.23\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at t=0.500 st=0.500 \ \text{s} is

vf=v0+at=(−14.0 m/s)+(−9.8 m/s2)(0.500 s)=−18.9 m/s  (Answer)\begin{align*} v_f & =v_0+at \\ &= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(0.500\:\text{s}\right) \\ & =-18.9\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

The displacement at t=1.00 st=1.00\ \text{s} is

Δy=v0t+12at2Δy=(−14.0 m/s)(1.00 s)+12(−9.8 m/s2)(1.00 s)2Δy=−18.9 m  (Answer)\begin{align*} \Delta y & =v_0t+\frac{1}{2}at^2 \\ \Delta y &=\left(-14.0\:\text{m/s}\right)\left(1.00\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(1.00\:\text{s}\right)^2 \\ \Delta y & =-18.9\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at t=1.00 st=1.00\ \text{s} is

vf=v0+at=(−14.0 m/s)+(−9.8 m/s2)(1.00 s)=−23.8 m/s  (Answer)\begin{align*} v_f & =v_0+at \\ &= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(1.00\:\text{s}\right) \\ & =-23.8\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

The displacement at t=1.50 st=1.50\ \text{s} is

Δy=v0t+12at2Δy=(−14.0 m/s)(1.50 s)+12(−9.8 m/s2)(1.50 s)2Δy=−32.0 m  (Answer)\begin{align*} \Delta y & =v_0t+\frac{1}{2}at^2 \\ \Delta y &=\left(-14.0\:\text{m/s}\right)\left(1.50\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(1.50\:\text{s}\right)^2 \\ \Delta y & =-32.0\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at t=1.50 st=1.50\ \text{s} is

vf=v0+at=(−14.0 m/s)+(−9.8 m/s2)(1.50 s)=−28.7 m/s  (Answer)\begin{align*} v_f & =v_0+at \\ &= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(1.50\:\text{s}\right) \\ & =-28.7\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part D

The displacement at t=2.00 st=2.00\ \text{s} is

Δy=v0t+12at2Δy=(−14.0 m/s)(2.00 s)+12(−9.8 m/s2)(2.00 s)2Δy=−47.6 m  (Answer)\begin{align*} \Delta y & =v_0t+\frac{1}{2}at^2 \\ \Delta y &=\left(-14.0\:\text{m/s}\right)\left(2.00\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(2.00\:\text{s}\right)^2 \\ \Delta y & =-47.6\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at t=2.00 st= 2.00 \ \text{s} is

vf=v0+at=(−14.0 m/s)+(−9.8 m/s2)(2.00 s)=−33.6 m/s  (Answer)\begin{align*} v_f & =v_0+at \\ &= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(2.00\:\text{s}\right) \\ & =-33.6\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part E

The displacement at t=2.50 st=2.50\ \text{s} is

Δy=v0t+12at2Δy=(−14.0 m/s)(2.50 s)+12(−9.8 m/s2)(2.50 s)2Δy=−65.6 m  (Answer)\begin{align*} \Delta y & =v_0t+\frac{1}{2}at^2 \\ \Delta y &=\left(-14.0\:\text{m/s}\right)\left(2.50\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(2.50\:\text{s}\right)^2 \\ \Delta y & =-65.6\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at t=2.50 st= 2.50 \ \text{s} is

vf=v0+at=(−14.0 m/s)+(−9.8 m/s2)(2.50 s)=−38.5 m/s  (Answer)\begin{align*} v_f & =v_0+at \\ &= \left(-14.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(2.50\:\text{s}\right) \\ & =-38.5\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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College Physics by Openstax Chapter 2 Problem 41

Calculate the displacement and velocity at times of (a) 0.500, (b) 1.00, (c) 1.50, and (d) 2.00 s for a ball thrown straight up with an initial velocity of 15.0 m/s. Take the point of release to be y0=0.

Solution:

The given known quantities are:a=−9.8 m/s2a=-9.8\:\text{m/s}^2; yo=0 my_o=0\:\text{m}; and voy=+15 m/sv_{oy}=+15\:\text{m/s}.

To compute for the displacement, we use the formula

Δy=voyt+12at2\Delta y=v_{oy}t+\frac{1}{2}at^2

and to compute for the final velocity, we use the formula

vfy=voy+atv_{fy}=v_{oy}+at

Part A

The displacement at t=0.500 st=0.500 \ \text{s} is

Δy=vot+12at2Δy=0 m+(15.0 m/s)(0.500 s)+12(−9.8 m/s2)(0.500 s)2Δy=6.28 m  (Answer)\begin{align*} \Delta y & =v_ot+\frac{1}{2}at^2 \\ \Delta y & =0\:\text{m}+\left(15.0\:\text{m/s}\right)\left(0.500\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(0.500\:\text{s}\right)^2 \\ \Delta y & =6.28\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at t=0.500 st=0.500 \ \text{s} is

vfy=voy+atvfy=(15.0 m/s)+(−9.8 m/s2)(0.500 s)vfy=10.1 m/s  (Answer)\begin{align*} v_{fy} & = v_{oy}+at \\ v_{fy} & =\left(15.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(0.500\:\text{s}\right) \\ v_{fy} & =10.1\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

The displacement at t=1.000 st=1.000 \ \text{s} is

Δy=vot+12at2Δy=0 m+(15.0 m/s)(1.000 s)+12(−9.8 m/s2)(1.000 s)2Δy=10.1 m  (Answer)\begin{align*} \Delta y & =v_ot+\frac{1}{2}at^2 \\ \Delta y & =0\:\text{m}+\left(15.0\:\text{m/s}\right)\left(1.000\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(1.000\:\text{s}\right)^2 \\ \Delta y & =10.1\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at t=1.000 st=1.000\ \text{s} is

vfy=voy+atvfy=(15.0 m/s)+(−9.8 m/s2)(1.000 s)vfy=5.20 m/s  (Answer)\begin{align*} v_{fy} & = v_{oy}+at \\ v_{fy} & =\left(15.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(1.000\:\text{s}\right) \\ v_{fy} & =5.20\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

The displacement at t=1.500 st=1.500\ \text{s} is

Δy=vot+12at2Δy=0 m+(15.0 m/s)(1.500 s)+12(−9.8 m/s2)(1.500 s)2Δy=11.5 m  (Answer)\begin{align*} \Delta y & =v_ot+\frac{1}{2}at^2 \\ \Delta y & =0\:\text{m}+\left(15.0\:\text{m/s}\right)\left(1.500\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(1.500\:\text{s}\right)^2 \\ \Delta y & =11.5\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at t=1.500 st=1.500\ \text{s} is

vfy=voy+atvfy=(15.0 m/s)+(−9.8 m/s2)(1.500 s)vfy=0.300 m/s  (Answer)\begin{align*} v_{fy} & = v_{oy}+at \\ v_{fy} & =\left(15.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(1.500\:\text{s}\right) \\ v_{fy} & =0.300\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part D

The displacement at t=2.000 st=2.000\ \text{s} is

Δy=vot+12at2Δy=0 m+(15.0 m/s)(2.000 s)+12(−9.8 m/s2)(2.000 s)2Δy=10.4 m  (Answer)\begin{align*} \Delta y & =v_ot+\frac{1}{2}at^2 \\ \Delta y & =0\:\text{m}+\left(15.0\:\text{m/s}\right)\left(2.000\:\text{s}\right)+\frac{1}{2}\left(-9.8\:\text{m/s}^2\right)\left(2.000\:\text{s}\right)^2 \\ \Delta y & =10.4\:\text{m}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

The velocity at t=2.000 st=2.000\ \text{s} is

vfy=voy+atvfy=(15.0 m/s)+(−9.8 m/s2)(2.000 s)vfy=−4.600 m/s  (Answer)\begin{align*} v_{fy} & = v_{oy}+at \\ v_{fy} & =\left(15.0\:\text{m/s}\right)+\left(-9.8\:\text{m/s}^2\right)\left(2.000\:\text{s}\right) \\ v_{fy} & =-4.600\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}
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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:v0=0 m/sv_0=0\:\text{m/s}; vf=65.0 m/sv_f=65.0\:\text{m/s}; and Δx=0.250 m\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

(vf)2=(v0)2+2aΔx\left(v_f\right)^2=\left(v_0\right)^2+2a\Delta x

We solve for acceleration in terms of the other variables.

a=(vf)2−(v0)22Δxa=\frac{\left(v_f\right)^2-\left(v_0\right)^2}{2\Delta x}

Substitute the given values

a=(vf)2−(v0)22Δxa=(65.0 m/s)2−(0 m/s)22(0.250 m)a=8450 m/s2  (Answer)\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

vf=v0+atv_f=v_0+at

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

t=vf−v0at=\frac{v_f-v_0}{a}

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

t=vf−v0at=65.0 m/s−0 m/s8450 m/s2t=7.6922 ×10−3 s  (Answer)\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 29

Freight trains can produce only relatively small accelerations and decelerations.

(a) What is the final velocity of a freight train that accelerates at a rate of 0.0500 m/s2 for 8.00 minutes, starting with an initial velocity of 4.00 m/s?

(b) If the train can slow down at a rate of 0.550 m/s2, how long will it take to come to a stop from this velocity?

(c) How far will it travel in each case?

Solution:

Part A

We are given the the following: a=0.0500 m/s2a=0.0500 \ \text{m/s}^2; t=8.00 minst=8.00 \ \text{mins}; and v0=4.00 m/sv_0=4.00 \ \text{m/s}.

The final velocity can be solved using the formula vf=v0+atv_f=v_0+at. We substitute the given values.

vf=v0+atvf=4.00 m/s+(0.0500 m/s2)(8.00 min×60 sec⁡1 min⁡)vf=28.0 m/s  (Answer)\begin{align*} v_f& = v_0+at \\ v_f & = 4.00\:\text{m/s}+\left(0.0500\:\text{m/s}^2\right)\left(8.00\:\text{min}\times \frac{60\:\sec }{1\:\min }\right) \\ v_f & = 28.0 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part B

Rearrange the equation we used in part (a) by solving in terms of tt, we have

t=vf−v0at=0 m/s−28 m/s−0.550 m/s2t=50.91 sec⁡  (Answer)\begin{align*} t & =\frac{{v_f}-v_0}{a} \\ t & = \frac{0\:\text{m/s}-28\:\text{m/s}}{-0.550\:\text{m/s}^2} \\ t & = 50.91\:\sec\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

Part C

The change in position for part (a), Δx \Delta x, or distance traveled is computed using the formula Δx=v0t+12at2 \Delta x=v_0 t+\frac{1}{2} at^2.

Δx=v0t+12at2Δx=(4.0 m/s)(480 s)+12(0.0500 m/s2)(480 s)2Δx=7680 m  (Answer)\begin{align*} \Delta x & =v_0 t+\frac{1}{2} at^2 \\ \Delta x & =\left(4.0\:\text{m/s}\right)\left(480\:\text{s}\right)+\frac{1}{2}\left(0.0500\:\text{m/s}^2\right)\left(480\:\text{s}\right)^2 \\ \Delta x & = 7680\:\text{m} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \end{align*}

For the situation in part (b), the distance traveled is computed using the formula Δx=vf2−v022a\Delta x=\frac{v_f^2-v_0^2}{2 a}.

Δx=(0 m/s)2−(28.0 m/s)22(−0.550 m/s2)Δx=712.73 m  (Answer)\begin{align*} \Delta x & =\frac{\left(0\:\text{m/s}\right)^2-\left(28.0\:\text{m/s}\right)^2}{2\left(-0.550\:\text{m/s}^2\right)} \\ \Delta x & =712.73\:\text{m} \ \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: v0=0 m/sv_0=0 \ \text{m/s}; vf=26.8 m/sv_f=26.8 \ \text{m/s}; and t=3.90 st=3.90\ \text{s}.

Part A

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

a‾=ΔvΔta‾=26.8 m/s−0 m/s3.90 sa‾=6.872 m/s2  (Answer)\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,

Δx=vavetΔx=(0 m/s+26.8 m/s2)(3.90 s)Δx=52.26 m  (Answer)\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 Δx=vavet\Delta x=v_{ave} t. That is,

Δx=vavetΔx=(8 m/s+40 m/s2)(3.33×10−2 s)Δx=0.7992 m  (Answer)\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: v0=0 m/sv_0=0\:\text{m/s}
Final Velocity: vf=30.0 cm/sv_f=30.0\:\text{cm/s}
Distance Traveled: x−x0=1.80 cmx-x_0=1.80\:\text{cm}

Part C

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

Δx=vavett=Δxvavet=1.80 cm(0 cm/s+30 cm/s)2t=0.12 s  (Answer)\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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College Physics by Openstax Chapter 2 Problem 25

At the end of a race, a runner decelerates from a velocity of 9.00 m/s at a rate of 2.00 m/s2.

(a) How far does she travel in the next 5.00 s?

(b) What is her final velocity?

(c) Evaluate the result. Does it make sense?

Solution:

We are given the following: v0=9.00 m/sv_0=9.00 \ \text{m/s}; and a=2.00 m/s2a=2.00 \ \text{m/s}^2.

Part A

For this part, we are given t=5.00 st=5.00 \ \text{s} and we shall use the formula x=x0+v0t+12at2 x=x_0+v_0 t+\frac{1}{2}at^2.

x=x0+v0t+12at2x=0 m+(9.00 m/s)(5.00 s)+12(−2.00 m/s2)(5.00 s)2x=20 meters  (Answer)\begin{align*} x & =x_0+v_0 t+\frac{1}{2}at^2 \\ x & =0\:\text{m}+\left(9.00\:\text{m/s}\right)\left(5.00\:\text{s}\right)+\frac{1}{2}\left(-2.00\:\text{m/s}^2\right)\left(5.00\:\text{s}\right)^2 \\ x & =20\:\text{meters} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}

Part B

The final velocity can be determined using the formula vf=v0+atv_f=v_0+at.

vf=v0+atvf=9.00 m/s+(−2.00 m/s2)(5.00 s)vf=−1 m/s  (Answer)\begin{align*} v_f & =v_0+at \\ v_f & =9.00\:\text{m/s}+\left(-2.00\:\text{m/s}^2\right)\left(5.00\:\text{s}\right) \\ v_f & =-1\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\ \end{align*}

Part C

The result says that the runner starts at the rate of 9 m/s and decelerates at 2 m/s2. After some time, the velocity is already negative. This does not make sense because if the velocity is negative, that means that the runner is already running backwards.

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