Chapter 4, Problem 77AP

(a)

To determine

The speed of the car at the edge of the cliff.

Answer to Problem 77AP

The speed of the car is $\text{20}\text{.0}\text{m/s}$.

Explanation of Solution

Draw the below figure as per given instruction.

The car has acceleration while it is on the slope and a different acceleration when it is falling. Take the motion apart into two different sections.

A chunk of motion is during the acceleration stays constant. The acceleration will change instantaneously at the brink of the cliff. The velocity and the position must be the same just before point $B$ and just after point $B$.

From point $A$ to point $B$ (along the incline), the car can be modeled as a particle under constant acceleration in one dimension, starting from rest $\left(v_i=0\right)$.

Write the expression for position along the incline plane as.

    $v_{{}_f}^2-v_i^2=2a\Delta x$                                                                                                (I)

Here, $v_f$ is final velocity, $v_i$ is initial velocity, $a$ is acceleration and $\Delta x$ is displacement.

Conclusion:

Substitute $0$ for $v_i$, $4.00{\text{ m/s}}^2$ for $a$ and $50.0\text{ m}$ for $\Delta x$ in equation (I).

    $\begin{array}{c}{v}_{{}_f}^2-0=2\left(4.00{\text{ m/s}}^2\right)\left(50.0\text{ m}\right)\\ =400.00\left({\text{m/s}}^2\right)\end{array}$

Solve and rearrange the above expression in terms of $v_f$ as.

    $\begin{array}{c}{v}_f=\sqrt{400.00}\text{ m/s}\\ \text{=20}\text{.0}\text{m/s}\end{array}$

Thus, the speed of the car is $\text{20}\text{.0}\text{m/s}$.

(b)

To determine

The elapsed time interval.

Answer to Problem 77AP

The time elapsed time interval is $5.00\text{ s}$.

Explanation of Solution

Use first equation of motion for the time interval.

Write the expression for motion in straight line as.

    $v_f=v_i+at$

Here, $t$ is the time interval.

Rearrange the above equation in terms of $t$ as.

    $t=\frac{v_f-v_i}{a}$                                                                                                    (II)

Conclusion:

Substitute $0$ for $v_i$, $4.00{\text{ m/s}}^2$ for $a$ and $20.0\text{ m/s}$ for $v_f$ in equation (II)

    $\begin{array}{c}t=\frac{20.0\text{ m/s}-0}{4.00{\text{ m/s}}^2}\\ =\frac{20.0\text{ m/s}}{4.00{\text{ m/s}}^2}\\ =5.00\text{ s}\end{array}$

Thus, the time elapsed time interval is $5.00\text{ s}$.

(c)

To determine

The velocity of the car.

Answer to Problem 77AP

The velocity of the car is $\left(16.0\widehat{i}-21.7\widehat{j}\right)\text{ m/s}$.

Explanation of Solution

Initial velocity is same as final velocity as horizontal acceleration is zero for uniform velocity. The velocity can be written as initial and final velocity component.

Write the expression for final velocity along horizontal as.

    $v_{xi}=v_i\cos \theta$                                                                                                (III)

Here, $v_{xi}$ is initial velocity component along horizontal and $\theta$ is inclination angle.

Write the expression for initial velocity along vertical as.

    $v_{yi}=-v_i\cos \theta$                                                                                              (IV)

Here, $v_{yi}$ is initial velocity component along vertical.

Write the expression for final velocity in $y$ direction as.

  $v_{yf}^2=v_{yi}^2+2a_y\Delta y$

Here, $v_{yf}$ is final velocity along vertical, $v_{yi}$ is initial velocity along vertical, $a_y$ is acceleration due to gravity and $\Delta y$ is vertical displacement in downwards direction.

Rearrange the above expression as final velocity as.

    $v_{yf}=-\sqrt{v_{yi}^2+2a_y\Delta y}$                                                                                    (V)

Write the expression for total velocity for the whole distance as.

    $v_f=v_{xi}\widehat{i}+v_{yf}\widehat{j}$                                                                                            (VI)

Here, $v_f$ is final velocity. $v_{xi}\widehat{i}$ is velocity along horizontal and $v_{yf}\widehat{j}$ is velocity along vertical.

Conclusion:

Substitute $20.0\text{ m/s}$ for $v_i$ and $37°$ for $\theta$ in equation (III)

    $\begin{array}{c}{v}_{xi}=20.0\text{ m/s}\left(\cos 37°\right)\\ =16.0\text{ m/s}\end{array}$

Substitute $20.0\text{ m/s}$ for $v_i$ and $37°$ for $\theta$ in equation (IV)

    $\begin{array}{c}{v}_{yi}=-20.0\text{ m/s}\left(\sin 37°\right)\\ =-12.0\text{ m/s}\end{array}$

Substitute $-12.0\text{ m/s}$ for $v_{yi}$, $-9.80{\text{ m/s}}^2$ for $a_y$ and $-30.0\text{ m}$ for $\Delta y$ in equation (V)

    $\begin{array}{c}{v}_{yf}=-\sqrt{{\left(-12.0\text{ m/s}\right)}^2+2\left(-9.80{\text{ m/s}}^2\right)\left(-30.0\text{ m}\right)}\\ =-\sqrt{144.0{\left(\text{m/s}\right)}^2+588{\left(\text{m/s}\right)}^2}\\ =-\sqrt{732{\left(\text{m/s}\right)}^2}\\ =-27.1\text{ m/s}\end{array}$

Substitute $-27.1\text{ m/s}$ for $v_{yf}$ and $16.0\text{ m/s}$ for $v_{xi}$ in equation (VI)

    $v_f=\left(16.0\widehat{i}-27.1\widehat{j}\right)\text{ m/s}$

Thus, the velocity of the car is $\left(16.0\widehat{i}-21.7\widehat{j}\right)\text{ m/s}$.

(d)

To determine

The total time interval the car is in motion.

Answer to Problem 77AP

The time interval for the total journey is $6.53\text{ s}$.

Explanation of Solution

Refer to drawn figure, the total time interval can be written as the sum of time required to cover distance $AB$ and the time required to cover distance $BC$.

Write the expression for total time interval as.

    $\Delta t=t_1+t_2$                                                                                                 (VIII)

Here, $\Delta t$ is time interval, $t_1$ is time for distance $AB$ and $t_2$ is time for distance $BC$.

Write the expression for time required to cover distance $BC$ as.

    $t_2=\frac{v_{yf}-v_{yi}}{a_y}$                                                                                                (IX)

Conclusion:

Substitute $-27.1\text{ m/s}$ for $v_{yf}$, $-12.0\text{ m/s}$ for $v_{yi}$ and $-9.80{\text{ m/s}}^2$ for $a_y$ in equation (IX)

    $\begin{array}{c}{t}_2=\frac{-27.1\text{ m/s}-\left(-12.0\text{ m/s}\right)}{-9.80{\text{ m/s}}^2}\\ =\frac{-15.1\text{ m/s}}{-9.80{\text{ m/s}}^2}\\ =1.53\text{ s}\end{array}$

Substitute $5.0\text{ s}$ for $t_1$ and $1.53\text{ s}$ for $t_2$ in equation (VIII)

    $\begin{array}{c}\Delta t=5.0\text{ s}+1.53\text{ s}\\ \text{=6}\text{.53 s}\end{array}$

Thus, the time interval for the total journey is $6.53\text{ s}$.

(e)

To determine

The position of the car.

Answer to Problem 77AP

The horizontal distance covered by the time $t_2$ is $24.5\widehat{i}\text{ m}$.

Explanation of Solution

Write the expression for horizontal distance in time $t_2$ as.

    $\Delta x=v_x\widehat{i}{t}_2$                                                                                                     (X)

Conclusion:

Substitute $16.0\text{ m/s}$ for $v_{xi}$ and $1.53\text{ s}$ for $t_2$ in equation (X)

    $\begin{array}{c}\Delta x=16.0\text{ m/s}\left(1.53\text{ s}\right)\\ =24.48\text{ m}\\ =\text{24}\text{.5}\widehat{i}\text{ m}\end{array}$

Thus, the horizontal distance covered by the time $t_2$ is $24.5\widehat{i}\text{ m}$.