Problem 13 - 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) 24.0 m/s, (b) 31.1 m/s.

Step 1: Identify the knowns and unknowns

We need to find (a) the initial horizontal velocity required for the car to clear the river and (b) the final speed of the car just before landing. The height difference between the starting and landing points is from 21.3 m to 1.8 m, with a horizontal distance of 48.0 m to cover.

Step 2: Establish equations of motion

For the horizontal motion: The car must travel 48.0 m. For vertical motion: solve for the time to fall from a height of 21.3 m to 1.8 m using the equation \[ y = v_{y0}t + \frac{1}{2}gt^2 \] where \( v_{y0} = 0 \), \( g = 9.8 \text{ m/s}^2 \), and \( y = 21.3 - 1.8 \). This gives the equation to solve for time \( t \).

Step 3: Solve for the time to land (t)

Substitute and solve the vertical equation:\[ 21.3 - 1.8 = \frac{1}{2}(9.8)t^2 \]\[ 19.5 = 4.9t^2 \]\[ t^2 = \frac{19.5}{4.9} \approx 3.98 \]\[ t \approx \sqrt{3.98} \approx 2.0 \text{ seconds} \]

Step 4: Calculate the initial horizontal velocity

Using the horizontal motion equation \( x = v_{x0} t \), where \( x = 48.0 \, \text{m} \), solve for \( v_{x0} \):\[ v_{x0} = \frac{48.0 \, \text{m}}{2.0 \, \text{s}} = 24.0 \, \text{m/s} \]

Step 5: Calculate the final speed upon landing

The final speed of the car just before landing can be found using its horizontal and vertical components of velocity. The horizontal velocity \( v_{x} \) remains \( 24.0 \, \text{m/s} \). The vertical velocity \( v_{y} \) when landing can be calculated using:\[ v_{y} = gt = 9.8 \, \text{m/s}^2 \times 2.0 \, \text{s} = 19.6 \, \text{m/s} \]Combine these into a total speed using\[ v = \sqrt{v_{x}^2 + v_{y}^2} = \sqrt{(24.0)^2 + (19.6)^2} \approx 31.1 \, \text{m/s} \]

Key Concepts

Equations of Motion

In physics, equations of motion are equations that describe the behavior of a moving object. These equations help us understand how things move over time. They are used to solve a variety of problems, like predicting the path of a car jumping over a river during a storm. Equations of motion typically come in handy when looking at the key variables that govern movement: displacement, velocity, acceleration, and time.

There are generally three primary equations of motion that are utilized to examine linear motion:

• The displacement equation: \( s = ut + \frac{1}{2}at^2 \), where \( s \) is displacement, \( u \) is initial velocity, \( a \) is acceleration, and \( t \) is time.
• The final velocity equation: \( v = u + at \), relating final velocity, initial velocity, acceleration, and time.
• The velocity-squared equation: \( v^2 = u^2 + 2as \), linking velocities, acceleration, and displacement.

In projectile motion, these equations help deduce how long it will take an object to hit the ground and how far it will travel horizontally. Understanding these equations is crucial for approaching most physics problem-solving scenarios.

Horizontal and Vertical Components

A projectile's motion can be understood by separating it into two orthogonal components: horizontal and vertical. By doing this, we can analyze each part independently, which simplifies calculations.

The horizontal component involves the horizontal distance or "range" an object travels. During projectile motion, the horizontal velocity \( v_{x0} \) remains constant because no external forces like air resistance or additional forces are considered. We use the formula \( x = v_{x0}t \) to compute how far the projectile travels horizontally in a given time \( t \).

The vertical component deals with how high or low the object travels relative to its starting point. It is influenced by gravity, making it accelerate downwards at \( 9.8 \text{ m/s}^2 \). The initial vertical velocity, often starting at zero, changes due to gravity as expressed by: \( v_{y} = v_{y0} + gt \). The displacement in vertical motion is given by \( y = v_{y0}t + \frac{1}{2}gt^2 \).

By combining findings from these sections, particularly in a physics problem-solving context, we can find the total motion behavior of the object, such as the car in the example.

Physics Problem Solving

Solving physics problems, like the scenario where a car needs to jump over a washed-out bridge, follows a structured method. This involves a step-by-step approach based on physics principles and equations.

First, identify knowns and unknowns. Spot the values given in the problem — like distances, heights, and time — and what the problem asks to find. This helps in forming equations using appropriate physics laws.

• For instance, understanding the heights the car starts and lands can dictate vertical motion equations.
• Recognizing the span of the river assists in calculating horizontal motion.

Next, apply the relevant equations of motion. This includes using known variables and maths to express the paths or velocity of the car.

• In our exercise, it involved determining the time the car takes to land using vertical motion equations, then using time to figure out horizontal reach and required speeds.

Finally, solve the equations. Do this to find unknown quantities, such as velocity just before landing, by breaking the problem into smaller, more manageable calculations. Mastering this systematic approach is key to tackling more complex kinematics questions.

Kinematics

Kinematics is the branch of physics that describes the motion of points, bodies, and systems of bodies without considering the forces that caused the motion. It’s all about the motion itself. In the case of projectile motion, kinematics helps in calculating trajectories and predicting where an object will land.

When analyzing kinematics, variables like displacement, velocity, and acceleration are central to building a comprehensive picture of motion.

• For motion in a straight line (linear motion), these include time-based calculations and distance covered.
• In two-dimensional motion, such as a projectile or a car jumping over a gap, both horizontal and vertical paths are analyzed, as these components together sketch out the trajectory seen in kinematics.

Applying kinematics in real-world scenarios, like predicting if a car can leap across a river, involves using its concepts and equations, focusing on time, speed, and distances traveled. Understanding kinematics enables students to solve such real-world physics puzzles by predicting if, where, and how an object in motion will perform given its initial conditions.