Chapter 3, Problem 83P

(a)

To determine

The reaction time.

Answer to Problem 83P

The reaction time is $\text{0}\text{.30 s}$

Explanation of Solution

Write the equation of motion.

  $y-y_0=v_0{}^2+\frac{1}{2}a{t}^2$        (I)

Here, $v_0$ is the final velocity, $t$ is the time, $a$ is the acceleration, and  $y-y_0$ is the fallen distance.

Conclusion:

Substitute $0$ for $v_0$, $g$ for $a$,in expression (I)

    $y–y_0 =–\frac{1}{2}g{t}^2$        (II)

Substitute $-0.45\text{m}$ for $y–y_0$, $\text{9}\text{.8}\text{m}\text{}\text{/}\text{}{\text{s}}^2$ for $g$,in expression (II) and solve for time,

  $\begin{array}{c}t=\sqrt{\frac{2(y-y_0)}{-g}}\\ =\sqrt{\frac{2(-0.45\text{m})}{\text{(}-\text{9}\text{.8}\text{m}\text{}\text{/}\text{}{\text{s}}^{\text{2}}\text{)}}}\\ =\text{0}\text{.30 s}\end{array}$

The reaction time is $\text{0}\text{.30 s}$

(b)

To determine

The distance for each time intervals.

Answer to Problem 83P

The distance for each time intervals are, $\text{for }t=\text{0}\text{.14 s, 10 cm}$, $\text{for }t=\text{0}\text{.16 s, 13 cm}$, $\text{for }t=\text{0}\text{.18 s, 16 cm}$, $\text{for }t=\text{0}\text{.20 s, 20 cm}$ and $\text{for }t=\text{0}\text{.22 s, 24 cm}$.

Explanation of Solution

Write the equation of motion.

  $y-y_0=v_0{}^2+\frac{1}{2}a{t}^2$        (I)

Here, $v_0$ is the final velocity, $t$ is the time, $a$ is the acceleration, and  $y-y_0$ is the fallen distance.

Conclusion:

Substitute $0$ for $v_0$, $g$ for $a$,in expression (I)

    $\begin{array}{l}y–y_0 =–\frac{1}{2}g{t}^2\\ y_0–y=4.9\times t^2\end{array}$        (II)

Substitute $0.14\sec$ for $t$, in expression (II) and solve for distance,

    $\begin{array}{c}{y}_0–y=4.9\times {\left(0.14\sec \right)}^2\\ =10\text{cm}\end{array}$

Substitute $0.16\sec$ for $t$, in expression (II) and solve for distance,

    $\begin{array}{c}{y}_0–y=4.9\times {\left(0.16\sec \right)}^2\\ =13\text{cm}\end{array}$

Substitute $0.18\sec$ for $t$, in expression (II) and solve for distance,

    $\begin{array}{c}{y}_0–y=4.9\times {\left(0.18\sec \right)}^2\\ =16\text{cm}\end{array}$

Substitute $0.20\sec$ for $t$, in expression (II) and solve for distance,

    $\begin{array}{c}{y}_0–y=4.9\times {\left(0.20\sec \right)}^2\\ =20\text{cm}\end{array}$

Substitute $0.22\sec$ for $t$, in expression (II) and solve for distance,

    $\begin{array}{c}{y}_0–y=4.9\times {\left(0.22\sec \right)}^2\\ =24\text{cm}\end{array}$

The distance for each time intervals are, $\text{for }t=\text{0}\text{.14 s, 10 cm}$, $\text{for }t=\text{0}\text{.16 s, 13 cm}$, $\text{for }t=\text{0}\text{.18 s, 16 cm}$, $\text{for }t=\text{0}\text{.20 s, 20 cm}$ and $\text{for }t=\text{0}\text{.22 s, 24 cm}$.