Chapter 4, Problem 4.77AP

A car is parked on a steep incline, making an angle of 37.0° below the horizontal and overlooking the ocean when its brakes fail and it begins to roll. Starting from rest at t = 0, the car rolls down the incline with a constant acceleration of 4.00 m/s², traveling 50.0 m to the edge of a vertical cliff. The cliff is 30.0 m above the ocean. Find (a) the speed of the car when it reaches the edge of the dill, (b) the time interval elapsed when it at rives there. (c) the velocity of the car when it lands in the ocean, (d) the total time interval the cat is in motion, and (e) the position of the car when it lands in the ocean, relative to the base of the cliff.

To determine

 The speed of car at the edge of cliff.

Answer to Problem 4.77AP

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

Explanation of Solution

Given info: The angle of the incline is ${37.0}^{\text{ο}}$ below the horizontal, the acceleration of car along he incline is $4.00\text{m}/{\text{s}}^2$ and the distance travelled by the car on the incline is $50.0\text{m}$. The height of cliff above the ocean is $30.0\text{m}$.

The value of acceleration due to gravity is $9.8\text{m}/{\text{s}}^2$.

The expression of kinematic equation of motion is,

$v^2-u^2=2as$

Here,

$v$ is the final velocity.

$u$ is the initial velocity.

$a$ is the acceleration.

$s$ is the distance covered.

Substitute $0$ for $u$, $4.00\text{m}/{\text{s}}^2$ for $a$ and $50.0\text{m}$ for $s$ in the above expression.

$\begin{array}{c}{v}^2-0=2\left(4.00\text{m}/{\text{s}}^2\right)\left(50.0\text{m}\right)\\ v=20\text{m}/\text{s}\end{array}$

Conclusion:

Therefore the speed of car at the edge of the cliff is $20\text{m}/\text{s}$.

To determine

 The time taken by the car to reach the edge of cliff.

Answer to Problem 4.77AP

 The time taken by the car to reach the edge of cliff is $5\text{s}$.

Explanation of Solution

Given info: The angle of the incline is ${37.0}^{\text{ο}}$ below the horizontal, the acceleration of car along he incline is $4.00\text{m}/{\text{s}}^2$ and the distance travelled by the car on the incline is $50.0\text{m}$. The height of cliff above the ocean is $30.0\text{m}$.

The expression of kinematic equation of motion is,

$s=ut+\frac{1}{2}a{t}^2$

Here,

$t$ is the time taken by the car.

Substitute $0$ for $u$, $4.00\text{m}/{\text{s}}^2$ for $a$ and $50.0\text{m}$ for $s$ in the above expression.

$\begin{array}{c}50.0\text{m}=\frac{1}{2}\left(4.00\text{m}/{\text{s}}^2\right)t^2\\ t=\sqrt{25{\text{s}}^2}\\ =5\text{s}\end{array}$

Conclusion:

Therefore the time taken by the car to reach the edge of cliff is $5\text{s}$.

To determine

 The velocity of the car when it lands in the ocean.

Answer to Problem 4.77AP

 The velocity of the car when it lands in the ocean is $16\text{m}/\text{s}\widehat{\text{i}}-27.1\text{m}/\text{s}\widehat{\text{j}}$.

Explanation of Solution

Given info: The angle of the incline is ${37.0}^{\text{ο}}$ below the horizontal, the acceleration of car along he incline is $4.00\text{m}/{\text{s}}^2$ and the distance travelled by the car on the incline is $50.0\text{m}$. The height of cliff above the ocean is $30.0\text{m}$.

The expression for the horizontal component of velocity at the edge of the cliff,

$u_{\text{h}}=v\cos \theta$

Substitute $-{37.0}^{\text{ο}}$ for $\theta$ and $20\text{m}/\text{s}$ for $v$ in the above expression.

$\begin{array}{c}{u}_{\text{h}}=\left(20\text{m}/\text{s}\right)\cos \left(-{37.0}^{\text{ο}}\right)\\ =16.0\text{m}/\text{s}\end{array}$

The horizontal component of velocity at the edge of the cliff is $16.0\text{m}/\text{s}$.

There is no acceleration of car at the edge of cliff, thus the value of horizontal component of velocity does not change.

The expression for the vertical component of velocity at the edge of the cliff,

$u_{\text{v}}=v\sin \theta$

Substitute $-{37.0}^{\text{ο}}$ for $\theta$ and $20\text{m}/\text{s}$ for $v$ in the above expression.

$\begin{array}{c}{u}_{\text{v}}=\left(20\text{m}/\text{s}\right)\sin \left(-{37.0}^{\text{ο}}\right)\\ =-12.0\text{m}/\text{s}\end{array}$

The vertical component of velocity at the edge of the cliff is $-12.0\text{m}/\text{s}$.

The expression of kinematic equation of motion is.

$v_{\text{v}}^2-u_{\text{v}}^2=2gh$

Here,

$g$ is the acceleration due to gravity.

$v_{\text{v}}$ is the vertical velocity when car lands in ocean.

$h$ is the height of the cliff above the ocean.

Rearrange the above expression for value of $v_{\text{v}}$.

$v_{\text{v}}^{}=-\sqrt{u_{\text{v}}^2+2gh}$

Substitute $-12.0\text{m}/\text{s}$ for $u_{\text{v}}$, $9.8\text{m}/{\text{s}}^2$ for $g$ and $30.0\text{m}$ for $h$ in the above expression.

$\begin{array}{c}{v}_{\text{v}}^{}=-\sqrt{{\left(-12.0\text{m}/\text{s}\right)}^2+2\left(9.8\text{m}/{\text{s}}^2\right)\left(30.0\text{m}\right)}\\ =-27.05\text{m}/\text{s}\\ \approx -27.1\text{m}/\text{s}\end{array}$

The expression for the velocity of the car, when it lands on the ocean is,

$v_{\text{c}}=u_{\text{h}}\widehat{\text{i}}+v_{\text{v}}^{}\widehat{\text{j}}$

Substitute $16\text{m}/\text{s}$ for $u_{\text{h}}$ and $-27.05\text{m}/\text{s}$ for $v_{\text{v}}^{}$ in the above expression.

$v_{\text{c}}=16\text{m}/\text{s}\widehat{\text{i}}-27.1\text{m}/\text{s}\widehat{\text{j}}$

Conclusion:

Therefore, the velocity of the car when it lands in the ocean is $16\text{m}/\text{s}\widehat{\text{i}}-27.1\text{m}/\text{s}\widehat{\text{j}}$.

To determine

 The total time interval of car in motion.

Answer to Problem 4.77AP

The total time interval of car in motion is $\text{6}\text{.53}\text{s}$.

Explanation of Solution

Given info: The angle of the incline is $37.0°$ below the horizontal, the acceleration of car along he incline is $4.00\text{m}/{\text{s}}^2$ and the distance travelled by the car on the incline is $50.0\text{m}$. The height of cliff above the ocean is $30.0\text{m}$.

The expression for kinematics equation of motion is,

$v_{\text{v}}=u_{\text{v}}-g{t}_{\text{f}}$

Here,

$t_{\text{f}}$ is the time period of fall of car.

Rearrange the above equation for the value of $t_{\text{f}}$.

$t_{\text{f}}=\frac{u_{\text{v}}-v_{\text{v}}}{g}$

Substitute $-12.0\text{m}/\text{s}$ for $u_{\text{v}}$, $9.8\text{m}/{\text{s}}^2$ for $g$ and $-27.0\text{m}/\text{s}$ for $v_{\text{v}}$ in the above expression.

$\begin{array}{c}{t}_{\text{f}}=\frac{-12.0\text{m}/\text{s}-\left(-27.0\text{m}/\text{s}\right)}{9.8\text{m}/{\text{s}}^2}\\ =1.53\text{s}\end{array}$

The time period of fall of car is $1.53\text{s}$.

The expression for the total time period of the motion of car is,

$t_{\text{t}}=t+t_{\text{f}}$

Substitute $5\text{s}$ for $t$ and $1.53\text{s}$ for $t_{\text{f}}$ in the above expression.

$\begin{array}{c}{t}_{\text{t}}=5\text{s}+1.53\text{s}\\ =\text{6}\text{.53}\text{s}\end{array}$

Conclusion:

Therefore, the total time interval of car in motion is $\text{6}\text{.53}\text{s}$.

To determine

 The position of the car at the time it lands in the ocean relative to the base of the cliff.

Answer to Problem 4.77AP

 The position of the car at the time it lands in the ocean relative to the base of the cliff is $24.5\widehat{\text{i}}\text{m}$.

Explanation of Solution

Given info: The angle of the incline is ${37.0}^{\text{ο}}$ below the horizontal, the acceleration of car along he incline is $4.00\text{m}/{\text{s}}^2$ and the distance travelled by the car on the incline is $50.0\text{m}$. The height of cliff above the ocean is $30.0\text{m}$.

The expression for horizontal distance travelled by the car during the fall from cliff is,

$d_{\text{h}}=u_{\text{h}}\times t_{\text{f}}$

Substitute $16\text{m}/\text{s}$ for $u_{\text{h}}$ and $1.53\text{s}$ for $t_{\text{f}}$ in the above expression.

$\begin{array}{c}{d}_{\text{h}}=16\text{m}/\text{s}\times 1.53\text{s}\\ =24.48\text{m}\\ \simeq 24.5\text{m}\end{array}$

The expression for position of car in vector form is,

$P=24.5\widehat{\text{i}}\text{m}$

Conclusion:

The position of the car at the time it lands in the ocean relative to the base of the cliff is $24.5\widehat{\text{i}}\text{m}$.