Q13E - University Physics with Modern Physics

Question

During a storm, a car traveling on a level horizontal road comes upon a bridge that has washed out. The driver must get to the other side, so he decides to try leaping the river with his car. The side of the road the car is on is 21.3 m above the river, while the opposite side is only 1.8 m above the river. The river itself is a raging torrent 48.0 m wide. (a) How fast should the car be traveling at the time it leaves the road in order just to clear the river and land safely on the opposite side? (b) What is the speed of the car just before it lands on the other side?

Step-by-Step Solution

Answer

a) The velocity of the car in the horizontal direction is 24.1 m/s .

b) The final velocity of the car is 31 m/s . 

Step 1: Identification of given data:

The given data can be listed below:

The width of the river is w = 48 m. 

The distance of car above the river is y = 21.3 m .

The distance of car on other side of river is $y_0=1.8 \mathrm{m}$ . 

Step 2: Concept/Significance of average velocity:

The overall displacement of an object divided by the total time it takes to move that thing is its average velocity.

Step 3: (a) Determination of fast should the car be traveling at the time:

The distance traveled by the car is,

$$\begin{gathered} y-y_0=21.3 \mathrm{m}-1.80 \mathrm{m} \\ =19.5 \mathrm{m} \end{gathered}$$ 

 

The time taken to cover the distance from the car is given by,

$$\begin{gathered} y-y_0=v_{0y}t+\frac{1}{2}{a}_y{t}^2 \\ t=\sqrt{\frac{2\left(y-y_0\right)}{a_y}} \end{gathered}$$ 

Here, $\left(y-y_0\right)$ is the distance covered in vertical direction, $y_{0y}$is the initial velocity in vertical direction, $a_y$is the acceleration in vertical direction.

 

Substitute al the values in the above expression.

$$\begin{gathered} t=\sqrt{\frac{2\left(19.5 \mathrm{m}\right)}{9.8 \mathrm{m}/{\mathrm{s}}^2}} \\ =1.99 \mathrm{s} \end{gathered}$$ 

 

The velocity of the car in the horizontal direction is given by,

$$\begin{gathered} \mathrm{x}-{\mathrm{x}}_0={\mathrm{v}}_{0\mathrm{x}}\mathrm{t}+\frac{1}{2}{\mathrm{a}}_{\mathrm{x}}{\mathrm{t}}^2 \\ \mathrm{x}-{\mathrm{x}}_0={\mathrm{v}}_{0\mathrm{x}}\mathrm{t} \\ {\mathrm{v}}_{0\mathrm{x}}=\frac{\mathrm{x}-{\mathrm{x}}_0}{\mathrm{t}} \end{gathered}$$ 

 

Here, $\mathrm{x}-{\mathrm{x}}_0$ is the horizontal distance travelled by the car,$a_x$is the acceleration in x-direction whose value is zero.

 

Substitute all the values in the above,

$$\begin{gathered} v_0=\frac{48 \mathrm{m}}{1.99 s} \\ =24.1 \mathrm{m}/\mathrm{s} \end{gathered}$$ 

 

Thus, the velocity of the car in the horizontal direction is 24.1 m/s  

Step 4: (b) Determination of the speed of the car just before it lands on the other side

The velocity in the vertical direction is given by,

${\mathrm{v}}_{\mathrm{y}}={\mathrm{v}}_{0\mathrm{y}}+{\mathrm{a}}_{\mathrm{y}}\mathrm{t}$ 

 

Here,${\mathrm{v}}_{0\mathrm{y}}$is the initial velocity in the vertical direction,${\mathrm{a}}_{\mathrm{y}}$is the acceleration in vertical direction, and t is the time taken.

 

Substitute all the values in the above,

 

$$\begin{gathered} {\mathrm{v}}_{\mathrm{y}}=\left(0 \mathrm{m}/\mathrm{s}\right)+\left(-9.8 \mathrm{m}/{\mathrm{s}}^2\right)\left(1.99 \mathrm{s}\right) \\ =-19.55 \mathrm{m}/\mathrm{s} \end{gathered}$$ 

 

The net velocity of the car is given by,

$\mathrm{v}=\sqrt{{\mathrm{v}}_{\mathrm{x}}^2+{\mathrm{v}}_{\mathrm{y}}^2}$ 

 

Substitute all the values in the above,

 $$\begin{gathered} v=\sqrt{{\left(24.1\mathrm{m}/\mathrm{s}\right)}^2+{\left(-19.55 \mathrm{m}/\mathrm{s}\right)}^2} \\ =\sqrt{961.086} \mathrm{m}/\mathrm{s} \\ =31 \mathrm{m}/\mathrm{s} \end{gathered}$$

 

Thus, the final velocity of the car is 31 m/s .