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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