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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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Problem 2-2: Distance and displacement of a given Path B

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PROBLEM:

Find the following for path B in the figure:

(a) The distance traveled.

(b) The magnitude of the displacement from start to finish.

(c) The displacement from start to finish.

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SOLUTION:

Part a

Based on the figure, B travels from 12 to 7. The distance traveled is 5 meters.

Part b

The magnitude of the displacement is 5 meters.

Part c

The displacement is calculated keeping in mind the sign. The motion started at 12 and ended at 7. Therefore, the displacement is

Δx=7 m−12 m=−5 meters  (Answer)\Delta x=7\:\text{m}-12\:\text{m}=-5\:\text{meters} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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