A helicopter's speed increases from $30 m/s$ to $40 m/s$ in 5 seconds. What is the acceleration of this helicopter?
Answer (2)
Define acceleration as the rate of change of velocity: a = t Δ v .
Calculate the change in velocity: Δ v = v f − v i = 40 s m − 30 s m = 10 s m .
Substitute the values into the acceleration formula: a = 5 s 10 s m = 2 s 2 m .
The acceleration of the helicopter is 2 s 2 m .
Explanation
Understanding the Problem We are given the initial velocity, final velocity, and the time interval over which the helicopter accelerates. We need to find the acceleration of the helicopter.
Defining Acceleration The acceleration is defined as the rate of change of velocity with respect to time. In other words, it's how much the velocity changes per unit of time. The formula for calculating acceleration when it is constant is: a = t Δ v = t v f − v i where:
a is the acceleration,
Δ v is the change in velocity,
v f is the final velocity,
v i is the initial velocity,
t is the time interval.
Calculating Acceleration We are given:
Initial velocity, v i = 30 s m
Final velocity, v f = 40 s m
Time interval, t = 5 s Now, we substitute these values into the formula: a = 5 s 40 s m − 30 s m a = 5 s 10 s m a = 2 s 2 m
Final Answer The acceleration of the helicopter is 2 s 2 m . This means that the helicopter's speed increases by 2 meters per second every second.
Examples
Understanding acceleration is crucial in many real-world scenarios, such as designing vehicles, analyzing the motion of objects, and understanding physics. For example, when designing a car, engineers need to calculate the acceleration required to reach a certain speed in a certain amount of time. Similarly, understanding the acceleration of a rocket is essential for space travel. In sports, understanding acceleration helps athletes improve their performance, such as a sprinter increasing their speed off the starting block.
The acceleration of the helicopter is 2 m/s 2 , calculated using the formula a = t Δ v where the change in velocity is 10 m/s over a time period of 5 seconds. This indicates the helicopter's speed increases by 2 meters per second every second. The calculations show the process clearly and accurately.
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