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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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The position-time (x−tx-t) graph represents the motion of an object over time. The slope of the tangent at any point on this graph gives the instantaneous velocity v=dxdtv = \frac{dx}{dt}. A straight line indicates constant velocity, while a curve indicates acceleration.

A position-time graph showing a parabolic curve indicating constant acceleration.
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In a velocity-time (v−tv-t) graph, the slope of the line represents the acceleration a=dvdta = \frac{dv}{dt}. A horizontal line represents zero acceleration (constant velocity), while a line with a constant non-zero slope represents uniform acceleration.

A velocity-time graph showing a straight line with a positive slope indicating uniform acceleration.
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The area under the velocity-time graph between two time intervals t1t_1 and t2t_2 gives the displacement SS of the object during that interval. Displacement is the algebraic sum of areas (considering signs), while distance is the sum of absolute magnitudes of areas.

Velocity-time graph where the area under the constant velocity line is shaded to represent displacement.
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An object at rest is represented by a horizontal line in an x−tx-t graph (v=0v = 0). In a v−tv-t graph, an object at rest or moving with constant velocity has a slope of zero (a=0a = 0).

📐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.

Problem 3:

A particle moves according to the velocity-time graph shown. Find the total distance covered and the displacement of the particle from t=0t = 0 to t=6 st = 6 \text{ s}.

Velocity-time graph with a positive peak at 2s and a negative peak at 5s.

Solution:

  1. Area of triangle from t=0t=0 to t=4t=4: A1=12×4×10=20 mA_1 = \frac{1}{2} \times 4 \times 10 = 20 \text{ m}.
  2. Area of triangle from t=4t=4 to t=6t=6: A2=12×2×(−10)=−10 mA_2 = \frac{1}{2} \times 2 \times (-10) = -10 \text{ m}.
  3. Total Displacement =A1+A2=20−10=10 m= A_1 + A_2 = 20 - 10 = 10 \text{ m}.
  4. Total Distance =∣A1∣+∣A2∣=20+10=30 m= |A_1| + |A_2| = 20 + 10 = 30 \text{ m}.

Explanation:

Displacement is the vector sum of areas under the v−tv-t graph, whereas distance is the scalar sum of the magnitudes of those areas.

Problem 4:

A body starts from the origin and its velocity increases linearly with time as v=2tv = 2t. Plot the x−tx-t graph for the first 4 s4 \text{ s} and find its position at t=4 st = 4 \text{ s}.

Position-time graph showing a parabola $x=t^2$ starting from the origin.

Solution:

Given v=dxdt=2tv = \frac{dx}{dt} = 2t. Integrating both sides: ∫0xdx=∫0t2t dt\int_{0}^{x} dx = \int_{0}^{t} 2t \, dt. x=t2x = t^2. At t=4 st = 4 \text{ s}, x=42=16 mx = 4^2 = 16 \text{ m}.

Explanation:

Since velocity is a linear function of time, position is a quadratic function of time, resulting in a parabolic x−tx-t graph.