91Ó°ÊÓ

Kinematic equations are fundamental when analyzing projectile motion. These equations relate displacement, velocity, acceleration, and time. They help us understand how objects move through space over time. In the context of the problem, the key kinematic equations used include:

• Vertical motion: \[ \Delta y = v_{y0}t + \frac{1}{2}gt^2 \]This equation calculates vertical displacement, where \( v_{y0} \) is the initial vertical velocity, \( t \) is time, and \( g \) is the acceleration due to gravity (approximately 9.81 m/s² downward).
• Horizontal motion: \[ \Delta x = v_{x} t \]This calculates horizontal displacement, where \( v_{x} \) is horizontal velocity.
• Velocity components: Finally, the overall speed just before landing, combining both vertical and horizontal velocities, is determined using:\[ v_{f} = \sqrt{v_{x}^2 + v_{y}^2} \]

These formulas give a comprehensive view of the car's path across the river, allowing us to calculate necessary speed and trajectory.

Vertical displacement is the distance moved by an object in the vertical direction. In the problem, the car starts from a height of 21.3 m and lands at 1.8 m. This gives it a vertical displacement of 19.5 m downward.

Since the car is dropping downwards under the influence of gravity, we use the kinematic equation:
\[ \Delta y = \frac{1}{2}gt^2 \]Here, the initial vertical velocity \( v_{y0} \) is zero because the car starts by moving horizontally.

• Given: \( \Delta y = 19.5 \ m \), and using \( g = 9.81 \ m/s^2 \).
• Rearranging for \( t \): \[ 19.5 = \frac{1}{2} \times 9.81 \times t^2 \]
• We solve for \( t \) and find it approximately equal to 1.994 seconds.

Understanding vertical displacement is crucial for calculating how long the car is in the air.

When discussing projectile motion, horizontal velocity is the speed at which an object travels parallel to the ground. It remains constant (ignoring air resistance) throughout the motion because no horizontal force acts on the object in flight.

For the car to clear a 61.0 m wide river, we use its horizontal velocity formula:\[ \Delta x = v_{x} t \]
This formula reveals the relation between distance \( \Delta x \), horizontal velocity \( v_{x} \), and time \( t \).

• In this example, \[ 61.0 = v_{x} \times 1.994 \]
• Solving for \( v_{x} \) gives around 30.6 m/s.

This represents how quickly the car must travel horizontally at the take-off to ensure it lands successfully on the opposite bank.

Understanding velocity components helps us see how velocity can be divided into horizontal and vertical directions. When the car lands, it has both horizontal and vertical velocity components.

The horizontal component \( v_{x} \) is 30.6 m/s, calculated from the need to cover the river's width. The vertical component \( v_{y} \) is due to gravity acting on the car:\[ v_{y} = gt = 9.81 \times 1.994 \approx 19.57 \ m/s \]These components combine into a final velocity just before the car lands. Using the Pythagorean theorem:\[ v_{f} = \sqrt{v_{x}^2 + v_{y}^2} = \sqrt{(30.6)^2 + (19.57)^2} \]\[ v_{f} = \sqrt{1320.2} \approx 36.3 \, \text{m/s} \]

• This resultant speed shows the overall magnitude of the car's velocity as it approaches the landing.

Breaking down velocity into components gives a clear view of its behavior in projectile motion.