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draw-velocity-time-graph-for-constant-acceleration

๐Ÿš€ Kinematics is the branch of mechanics that deals with the motion of objects without considering the forces that cause the motion. One important aspect of kinematics is uniformly accelerated motion, which occurs when an object moves in a straight line with a constant acceleration. In this scenario, the velocity of the object changes at a constant rate over time. The velocity-time graph for an object undergoing constant acceleration is a straight line, where the slope of the line represents the acceleration. The area under the velocity-time graph gives the displacement of the object during that time interval.

Theory Explanation

Understanding Uniformly Accelerated Motion

In uniformly accelerated motion, the acceleration (a) is constant. This means that the change in velocity (ฮ”v) over a time interval (ฮ”t) is the same throughout the motion. The relationship can be expressed as: a = ฮ”v / ฮ”t. If an object starts from an initial velocity (u) and accelerates at a constant rate (a) for a time (t), the final velocity (v) can be calculated using the equation: v = u + at.

\[ v = u + at \]
Velocity-Time Graph for Constant Acceleration

The velocity-time graph for an object with constant acceleration is a straight line. The slope of this line represents the acceleration. If the object starts from rest (u = 0), the graph will start from the origin (0,0). The area under the line between two points in time gives the displacement of the object during that time interval. For example, if the line goes from (0,0) to (t,v), the area under the graph (which is a triangle) can be calculated as: Area = 0.5 * base * height = 0.5 * t * v.

\[ Area = \frac{1}{2} \times t \times v \]

Key Points

  • ๐ŸŽฏ Kinematics deals with motion without considering forces.
  • ๐ŸŽฏ Uniformly accelerated motion has constant acceleration.
  • ๐ŸŽฏ The velocity-time graph for constant acceleration is a straight line.
  • ๐ŸŽฏ The slope of the velocity-time graph represents acceleration.
  • ๐ŸŽฏ The area under the velocity-time graph represents displacement.

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Examples:💡

An object starts from rest and accelerates at a rate of 2 m/sยฒ for 5 seconds. Find the final velocity and the displacement during this time.

Solution:

Step 1: Given initial velocity (u) = 0 m/s, acceleration (a) = 2 m/sยฒ, and time (t) = 5 s. Use the formula v = u + at to find the final velocity.

\[ v = 0 + (2)(5) = 10 m/s \]

Step 2: To find the displacement, use the formula: Displacement = Area under the velocity-time graph. The area of the triangle is given by: Area = 0.5 * base * height = 0.5 * t * v.

\[ Displacement = 0.5 * 5 * 10 = 25 m \]

A car moving with an initial velocity of 10 m/s accelerates at 3 m/sยฒ for 4 seconds. Calculate the final velocity and the distance covered during this time.

Solution:

Step 1: Initial velocity (u) = 10 m/s, acceleration (a) = 3 m/sยฒ, time (t) = 4 s. Use v = u + at to find the final velocity.

\[ v = 10 + (3)(4) = 22 m/s \]

Step 2: To find the distance covered, use the area under the velocity-time graph. The area of the trapezium can be calculated as: Area = 0.5 * (u + v) * t.

\[ Distance = 0.5 * (10 + 22) * 4 = 64 m \]

Common Mistakes

  • Mistake: Confusing acceleration with velocity; students often think that acceleration is the same as the change in velocity.

    Correction: Remember that acceleration is the rate of change of velocity, not the velocity itself.

  • Mistake: Incorrectly calculating the area under the velocity-time graph; students may forget that the area represents displacement.

    Correction: Always remember that the area under the graph gives the displacement, and use the correct formula for the shape of the area.

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