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Motion in a Straight Line - Position-Time and Velocity-Time Graphs

Grade 11CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Position-Time Graph (x−tx-t Graph): This graph represents the change in position of an object with respect to time. The slope of the tangent at any point on the x−tx-t graph gives the instantaneous velocity v=dxdtv = \frac{dx}{dt}.

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Types of x−tx-t Graphs: A horizontal line parallel to the time axis indicates the object is at rest (v=0v = 0). A straight line passing through the origin indicates uniform motion (constant velocity). A curved line (parabola) indicates non-uniform motion or accelerated motion.

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Velocity-Time Graph (v−tv-t Graph): This graph represents the change in velocity of an object with respect to time. The slope of the v−tv-t graph represents the instantaneous acceleration a=dvdta = \frac{dv}{dt}.

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Area under v−tv-t Graph: The area enclosed by the velocity-time graph and the time axis between t1t_1 and t2t_2 represents the displacement Δx\Delta x of the object during that time interval.

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Uniform Acceleration: In a v−tv-t graph, uniform acceleration is represented by a straight line with a constant slope. If the slope is positive, the object is accelerating; if the slope is negative, it is decelerating (retardation).

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Nature of Slopes: For an x−tx-t graph, a constant slope means constant velocity. For a v−tv-t graph, a constant slope means constant acceleration.

📐Formulae

Instantaneous Velocity: v=dxdt=tan⁡θ (Slope of x−t graph)\text{Instantaneous Velocity: } v = \frac{dx}{dt} = \tan \theta \text{ (Slope of } x-t \text{ graph)}

Instantaneous Acceleration: a=dvdt=tan⁡ϕ (Slope of v−t graph)\text{Instantaneous Acceleration: } a = \frac{dv}{dt} = \tan \phi \text{ (Slope of } v-t \text{ graph)}

Displacement: S=∫t1t2v dt=Area under v−t graph\text{Displacement: } S = \int_{t_1}^{t_2} v \, dt = \text{Area under } v-t \text{ graph}

Average Velocity: vavg=x2−x1t2−t1\text{Average Velocity: } v_{avg} = \frac{x_2 - x_1}{t_2 - t_1}

Distance: Total area under v−t graph (treating negative areas as positive)\text{Distance: } \text{Total area under } v-t \text{ graph (treating negative areas as positive)}

💡Examples

Problem 1:

A car starts from rest and accelerates uniformly to a velocity of 20 m/s20 \text{ m/s} in 10 s10 \text{ s}. It then moves with this constant velocity for 20 s20 \text{ s} and finally comes to rest in 5 s5 \text{ s} with uniform retardation. Calculate the total displacement using a v−tv-t graph.

Solution:

The motion is divided into three parts:

  1. Acceleration phase (0 to 10s): Area of triangle A1=12×10×20=100 mA_1 = \frac{1}{2} \times 10 \times 20 = 100 \text{ m}.
  2. Constant velocity phase (10 to 30s): Area of rectangle A2=20×20=400 mA_2 = 20 \times 20 = 400 \text{ m}.
  3. Retardation phase (30 to 35s): Area of triangle A3=12×5×20=50 mA_3 = \frac{1}{2} \times 5 \times 20 = 50 \text{ m}. Total Displacement S=A1+A2+A3=100+400+50=550 mS = A_1 + A_2 + A_3 = 100 + 400 + 50 = 550 \text{ m}.

Explanation:

Displacement is found by calculating the total area under the v−tv-t graph. The graph forms a trapezium, and the area is the sum of the areas of the geometric shapes formed.

Problem 2:

The x−tx-t graph for a particle is a parabola given by x=2t2x = 2t^2. Find the velocity of the particle at t=3 st = 3 \text{ s}.

Solution:

Given x=2t2x = 2t^2. Velocity v=dxdt=ddt(2t2)=4tv = \frac{dx}{dt} = \frac{d}{dt}(2t^2) = 4t. At t=3 st = 3 \text{ s}, v=4×3=12 m/sv = 4 \times 3 = 12 \text{ m/s}.

Explanation:

The velocity is the slope of the position-time graph. By differentiating the position function with respect to time, we obtain the instantaneous velocity at any given time tt.

Position-Time and Velocity-Time Graphs Revision - Class 11 Physics CBSE