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Motion is the change in position of an object with respect to time. To understand motion clearly, we first need to distinguish between some fundamental terms:
Understanding these concepts helps us analyze how objects move. The equations of motion are mathematical tools that describe the relationship between displacement, velocity, acceleration, and time when an object moves with uniform acceleration (constant acceleration).
These equations are essential for solving many practical problems, such as calculating the distance a vehicle covers while accelerating, or the time it takes for a stone to fall from a height.
When an object moves in a straight line with constant acceleration, the following variables are used:
By definition, acceleration is the rate of change of velocity:
\[a = \frac{v - u}{t}\]
Rearranging, we get:
\[v = u + at\]
This equation tells us the final velocity of an object after time t when it accelerates uniformly.
Displacement is the area under the velocity-time graph. Since velocity changes uniformly from \( u \) to \( v \), the average velocity is:
\[\text{Average velocity} = \frac{u + v}{2}\]
Displacement is then:
\[s = \text{average velocity} \times t = \frac{u + v}{2} \times t\]
Using the first equation \( v = u + at \), substitute \( v \):
\[s = \frac{u + (u + at)}{2} \times t = \left(u + \frac{at}{2}\right) t = ut + \frac{1}{2} at^2\]
From the first equation, \( v = u + at \), we can write:
\[t = \frac{v - u}{a}\]
Substitute \( t \) into the second equation:
\[s = ut + \frac{1}{2} at^2 = u \times \frac{v - u}{a} + \frac{1}{2} a \left(\frac{v - u}{a}\right)^2\]
Simplify:
\[s = \frac{u(v - u)}{a} + \frac{1}{2} a \times \frac{(v - u)^2}{a^2} = \frac{u(v - u)}{a} + \frac{(v - u)^2}{2a}\]
Multiply both sides by \( 2a \):
\[2as = 2u(v - u) + (v - u)^2\]
Expand the right side:
\[2as = 2uv - 2u^2 + v^2 - 2uv + u^2 = v^2 - u^2\]
Rearranged:
\[v^2 = u^2 + 2as\]
The velocity-time graph is a straight line with slope equal to acceleration. The area under this graph gives displacement \( s \). The displacement-time graph is a curve, showing increasing displacement with time under acceleration.
Step 1: Identify known values:
Step 2: Use the first equation of motion:
\[ v = u + at = 0 + (2)(5) = 10 \, m/s \]
Answer: The final velocity of the car is \( 10 \, m/s \).
Step 1: Known values:
Step 2: Use the second equation of motion:
\[ s = ut + \frac{1}{2} at^2 = (3)(4) + \frac{1}{2} (1.5)(4)^2 \]
\[ s = 12 + 0.75 \times 16 = 12 + 12 = 24 \, m \]
Answer: The runner covers a displacement of \( 24 \, m \) in 4 seconds.
Step 1: Known values:
Step 2: Use the third equation of motion:
\[ v^2 = u^2 + 2as \]
Rearranged to find acceleration \( a \):
\[ a = \frac{v^2 - u^2}{2s} = \frac{20^2 - 10^2}{2 \times 75} = \frac{400 - 100}{150} = \frac{300}{150} = 2 \, m/s^2 \]
Answer: The acceleration is \( 2 \, m/s^2 \).
Step 1: Known values:
Step 2: Find deceleration \( a \) using first equation:
\[ v = u + at \implies 0 = 18 + a \times 6 \implies a = -\frac{18}{6} = -3 \, m/s^2 \]
The negative sign indicates deceleration.
Step 3: Find stopping distance \( s \) using second equation:
\[ s = ut + \frac{1}{2} at^2 = 18 \times 6 + \frac{1}{2} \times (-3) \times 6^2 = 108 - 54 = 54 \, m \]
Answer: The deceleration is \( 3 \, m/s^2 \) and the stopping distance is \( 54 \, m \).
Step 1: Known values:
Step 2: Find time \( t \) using second equation:
\[ s = ut + \frac{1}{2} at^2 \implies 45 = 0 + \frac{1}{2} \times 9.8 \times t^2 \implies 45 = 4.9 t^2 \]
\[ t^2 = \frac{45}{4.9} \approx 9.18 \implies t = \sqrt{9.18} \approx 3.03 \, s \]
Step 3: Find velocity just before impact using first equation:
\[ v = u + at = 0 + 9.8 \times 3.03 = 29.7 \, m/s \]
Answer: The stone takes approximately \( 3.03 \) seconds to reach the ground and its velocity just before impact is \( 29.7 \, m/s \) downward.
When to use: Deciding which equation to apply based on given data.
When to use: When time and acceleration are known, and velocity is needed.
When to use: Finding final velocity from displacement and acceleration.
When to use: Always, especially in competitive exams.
When to use: When the object is slowing down.
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