Chapter 2, Problem 81CP

(a)

To determine

The distance between the nose of car and the south edge when the car stops.

Answer to Problem 81CP

The distance between the nose of car and the south edge when the car stops is $\underset{\_}{35.9\text{m}}$.

Explanation of Solution

Write the expression for the final position of the nose of the car.

    $x=x_0+v_{0}t+\frac{1}{2}a{t}^2$                                                                                         (I)

Here, $x$ is the final position of the car, $x_0$ is the initial position of car, $v_0$ is the initial velocity of car as it enters the intersection, $t$ is the time, and $a$ is the acceleration of car.

The initial position of the car is zero. Put $0\text{m}$ for $x_0$ in expression (I) and solve for $v_0$.

    $\begin{array}{c}x=v_{0}t+\frac{1}{2}a{t}^2\\ v_0=\frac{x-\frac{1}{2}a{t}^2}{t}\end{array}$                                                                                             (II)

Write the expression for the final velocity of the car.

    $v^2=v_0^2+2as$                                                                                                 (III)

Here, $v$ is the final velocity of the car, and $s$ is the displacement of the car.

Rearrange expression (III) to find $s$.

    $s=\frac{v^2-v_0^2}{2a}$                                                                                                            (IV)

Conclusion:

Substitute $28.0\text{m}$ for $x$, $3.10\text{s}$ for $t$, $-2.10{\text{m/s}}^2$ for $a$ in equation (II) to find $v_0$.

    $\begin{array}{c}{v}_0=\frac{28.0\text{m}-\frac{1}{2}\left(-2.10{\text{m/s}}^2\right){\left(3.10\text{s}\right)}^2}{\left(3.10\text{s}\right)}\\ =12.3\text{m/s}\end{array}$

Since the blue car stops at the intersection, final velocity is zero.

Substitute $12.3\text{m/s}$ for $v_0$ and $0\text{m/s}$ for $v$ in expression (IV) to find $s$.

    $\begin{array}{c}s=\frac{{\left(0\text{m/s}\right)}^2-{\left(12.3\text{m/s}\right)}^2}{2\left(-2.10{\text{m/s}}^2\right)}\\ =35.9\text{m}\end{array}$

Therefore, the distance between the nose of car and the south edge when the car stops is $\underset{\_}{35.9\text{m}}$.

(b)

To determine

The time interval in which the car is in the boundaries of intersection.

Answer to Problem 81CP

The time in which the car is in the boundaries of intersection is $\underset{\_}{4.04\text{s}}$.

Explanation of Solution

The time for which the car is in the intersection is the time between the entering of nose and exiting of tail from the intersection. Thus the total distance travelled by the car between the intersections is equal to the sum of length of car and length of the intersection path. Thus the change in position of nose of the car is equal to $4.52\text{m}+28.0\text{m}=32.52\text{m}$. Let at time $0\text{s}$, the velocity of car is $12.3\text{m/s}$ also the car starts from origin.

Write the expression for the final position of car at time $t$ from expression (I).

    $x=x_0+v_{0}t+\frac{1}{2}a{t}^2$

The car starts from origin, so $x_0$ is zero. Put $0\text{m}$ for $x_0$ in above expression and solve for $t$.

    $\begin{array}{c}x=v_{0}t+\frac{1}{2}a{t}^2\\ v_{0}t+\frac{1}{2}a{t}^2-x=0\end{array}$                                                                                (V)

Expression (V) is a quadratic equation.

Write the general expression for a quadratic equation in terms of $t$.

    $a{t}^2+bt+c=0$                                                                                                   (VI)

Here, $a,b\text{and }c$ are constants.

Write the expression to find the solution for quadratic equation (VI).

    $t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$                                                                                               (VII)

Conclusion:

Substitute $12.3\text{m/s}$ for $v_0$, $-2.10{\text{m/s}}^2$ for $a$, and $32.52\text{m}$ for $x$ in expression (V).

    $\begin{array}{c}\left(12.3\text{m/s}\right)t+\frac{1}{2}\left(-2.10{\text{m/s}}^2\right)t^2-32.52\text{m}=0\\ -1.05t^2+12.3t-32.52=0\end{array}$

Compare the above quadratic expression with (VI) to obtain the values of constants.

    $\begin{array}{l}a=-1.05\\ b=12.3\\ c=-32.52\end{array}$

Substitute $-1.05$ for $a$, $12.3$ for $b$, and $-32.52$ for $c$ in expression (VII) to find $t$.

    $\begin{array}{c}t=\frac{-12.3\pm \sqrt{{\left(12.3\right)}^2-4\left(-1.05\right)\left(-32.52\right)}}{2\left(-32.52\right)}\\ =4.04\text{and 7}\text{.76}\end{array}$

Thus the time values are $4.04\text{s}$ and $7.66\text{s}$. Out time must be the smallest value. So the answer is $4.04\text{s}$.

Therefore, the time in which the car is in the boundaries of intersection is $\underset{\_}{4.04\text{s}}$.

(c)

To determine

The minimum distance from the near edge of intersection where the red car can start its motion after the complete leaving of blue car.

Answer to Problem 81CP

The minimum distance from the near edge of intersection where the red car can starts its motion after the complete leaving of blue car is $\underset{\_}{45.8\text{m}}$.

Explanation of Solution

The nose of the blue car enters the intersection at $0\text{s}$. At $4.04\text{s}$, the tail of blue car leaves the intersection. Therefore the nose of red car must enter at $4.04\text{s}$ to find the minimum distance at which the red car is to be started.

Again use expression (I) to find the distance between near edge and nose of car.

    $x=x_0+v_{0}t+\frac{1}{2}a{t}^2$

Conclusion:

Substitute $0\text{m}$ for $x_0$, $0\text{m/s}$ for $v_0$, $5.60\text{m/s}$ for $a$, and $4.04\text{s}$ for $t$ in above equation to find $x$.

    $\begin{array}{c}x=\left(0\text{m}\right)+\left(0\text{m/s}\right)\left(4.04\text{s}\right)+\frac{1}{2}\left(5.60{\text{m/s}}^2\right){\left(4.04\text{s}\right)}^2\\ =45.8\text{m}\end{array}$

Therefore, the minimum distance from the near edge of intersection where the red car can starts its motion after the complete leaving of blue car is $\underset{\_}{45.8\text{m}}$.

(d)

To determine

The speed of red car when it enters the intersection.

Answer to Problem 81CP

The speed of red car when it enters the intersection is $\underset{\_}{22.6\text{m/s}}$.

Explanation of Solution

Write the expression to find the velocity of red car.

    $v=v_0+at$                                                                                                          (VIII)

Conclusion:

Substitute $0\text{m/s}$ for $v_0$, $5.60{\text{m/s}}^2$ for $a$, and $4.04\text{s}$ for $t$ in expression (VIII) to find $v$.

    $\begin{array}{c}v=0\text{m/s}+\left(5.60{\text{m/s}}^2\right)\left(4.04\text{s}\right)\\ =22.6\text{m/s}\end{array}$

Therefore, the speed of red car when it enters the intersection is $\underset{\_}{22.6\text{m/s}}$.