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When you watch a car speeding up on a highway or a cyclist slowing down before a turn, you are observing changes in their motion. But what exactly describes this change? This is where the concept of acceleration comes in. Acceleration is the measure of how quickly an object's velocity changes over time.
Velocity itself is a vector quantity, meaning it has both magnitude (speed) and direction. Acceleration, therefore, is also a vector and can indicate speeding up, slowing down, or changing direction.
When an object slows down, this is often called deceleration. Deceleration is simply acceleration with a negative value relative to the direction of motion. Understanding acceleration and deceleration is crucial for analyzing everyday motions, from vehicles on Indian roads to sports players on the field.
Acceleration is defined as the rate of change of velocity with respect to time. Mathematically, if an object's velocity changes from an initial velocity \( u \) to a final velocity \( v \) in a time interval \( t \), then the acceleration \( a \) is given by:
The unit of acceleration is meters per second squared (m/s²), which means the velocity changes by so many meters per second every second.
Positive acceleration means the velocity is increasing in the direction of motion, while negative acceleration (often called deceleration) means the velocity is decreasing.
Acceleration can be classified based on how it changes over time:
We also distinguish between:
Step 1: Identify given values: \( u = 0 \, m/s \), \( v = 20 \, m/s \), \( t = 5 \, s \).
Step 2: Use the acceleration formula:
\[ a = \frac{v - u}{t} = \frac{20 - 0}{5} = \frac{20}{5} = 4 \, m/s^2 \]
Answer: The car's acceleration is \( 4 \, m/s^2 \).
Step 1: Given: \( u = 30 \, m/s \), \( v = 0 \, m/s \), \( t = 10 \, s \).
Step 2: Apply the formula:
\[ a = \frac{v - u}{t} = \frac{0 - 30}{10} = \frac{-30}{10} = -3 \, m/s^2 \]
Step 3: The negative sign indicates deceleration.
Answer: The bus decelerates at \( 3 \, m/s^2 \).
Step 1: Identify velocities and times: \( v_1 = 12 \, m/s \) at \( t_1 = 2 \, s \), \( v_2 = 8 \, m/s \) at \( t_2 = 6 \, s \).
Step 2: Calculate time interval: \( \Delta t = 6 - 2 = 4 \, s \).
Step 3: Calculate change in velocity: \( \Delta v = 8 - 12 = -4 \, m/s \).
Step 4: Calculate average acceleration:
\[ a = \frac{\Delta v}{\Delta t} = \frac{-4}{4} = -1 \, m/s^2 \]
Answer: The average acceleration between 2 s and 6 s is \( -1 \, m/s^2 \), indicating deceleration.
Step 1: Given: \( u = 5 \, m/s \), \( v = 15 \, m/s \), \( t = 4 \, s \).
Step 2: Calculate acceleration using \( a = \frac{v - u}{t} \):
\[ a = \frac{15 - 5}{4} = \frac{10}{4} = 2.5 \, m/s^2 \]
Step 3: Use displacement formula:
\[ s = ut + \frac{1}{2} a t^2 = 5 \times 4 + \frac{1}{2} \times 2.5 \times 4^2 \]
\[ s = 20 + 0.5 \times 2.5 \times 16 = 20 + 20 = 40 \, m \]
Answer: The runner covers 40 meters during acceleration.
Step 1: Given: \( u = 0 \, m/s \), \( a = 2 \, m/s^2 \), \( s = 100 \, m \).
Step 2: Use the equation without time:
\[ v^2 = u^2 + 2as = 0 + 2 \times 2 \times 100 = 400 \]
Step 3: Calculate final velocity:
\[ v = \sqrt{400} = 20 \, m/s \]
Answer: The bike's final velocity is \( 20 \, m/s \).
When to use: Distinguishing between acceleration and deceleration problems.
When to use: Graph-based questions in exams.
When to use: Problems where acceleration is constant.
When to use: During formula application and unit conversions.
When to use: When the object is slowing down.
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