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Motion - Plot and interpret position-time and velocity-time graphs for uniform motion

Grade 9CBSE

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

🔑Concepts

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A position-time graph (or distance-time graph) for an object in uniform motion is always a straight line passing through the origin if the object starts from rest. The slope of this line represents the speed (or velocity) of the object, given by slope=change in positionchange in time\text{slope} = \frac{\text{change in position}}{\text{change in time}}.

A straight line graph showing position increasing linearly with time, indicating uniform velocity.
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In uniform motion, the velocity remains constant over time. Therefore, the velocity-time graph is a horizontal straight line parallel to the time axis (x-axis). The slope of this graph is zero, indicating that the acceleration is zero (a=0a = 0).

A horizontal line graph showing velocity remaining constant at 5 m/s as time increases.
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The area under a velocity-time graph represents the displacement of the object. For uniform motion, this area forms a rectangle. The area is calculated as Area=velocity×time interval\text{Area} = \text{velocity} \times \text{time interval}, which equals the magnitude of displacement ss.

A velocity-time graph where the area under the horizontal line between two time points is shaded to represent displacement.
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The steepness of the slope in a position-time graph indicates the magnitude of the velocity. A steeper line represents a higher constant velocity, while a less steep line represents a lower constant velocity.

📐Formulae

v=stv = \frac{s}{t}

Average Speed=Total DistanceTotal Time\text{Average Speed} = \frac{\text{Total Distance}}{\text{Total Time}}

Average Velocity=Total DisplacementTotal Time\text{Average Velocity} = \frac{\text{Total Displacement}}{\text{Total Time}}

a=v−uta = \frac{v - u}{t}

v=u+atv = u + at

s=ut+12at2s = ut + \frac{1}{2}at^2

v2=u2+2asv^2 = u^2 + 2as

💡Examples

Problem 1:

An object travels 16 m16\text{ m} in 4 s4\text{ s} and then another 16 m16\text{ m} in 2 s2\text{ s}. What is the average speed of the object?

Solution:

Total distance traveled by the object =16 m+16 m=32 m= 16\text{ m} + 16\text{ m} = 32\text{ m}. Total time taken =4 s+2 s=6 s= 4\text{ s} + 2\text{ s} = 6\text{ s}. Average speed =Total distanceTotal time=32 m6 s=5.33 m/s= \frac{\text{Total distance}}{\text{Total time}} = \frac{32\text{ m}}{6\text{ s}} = 5.33\text{ m/s}.

Explanation:

Since the object covers the same distance in different time intervals, this is an example of non-uniform motion. Average speed is used to describe the overall rate of motion.

Problem 2:

A bus decreases its speed from 80 km/h80\text{ km/h} to 60 km/h60\text{ km/h} in 5 s5\text{ s}. Find the acceleration of the bus.

Solution:

Initial velocity u=80 km/h=80×518 m/s=22.22 m/su = 80\text{ km/h} = 80 \times \frac{5}{18}\text{ m/s} = 22.22\text{ m/s}. Final velocity v=60 km/h=60×518 m/s=16.67 m/sv = 60\text{ km/h} = 60 \times \frac{5}{18}\text{ m/s} = 16.67\text{ m/s}. Time t=5 st = 5\text{ s}. Acceleration a=v−ut=16.67−22.225=−1.11 m/s2a = \frac{v - u}{t} = \frac{16.67 - 22.22}{5} = -1.11\text{ m/s}^2.

Explanation:

The negative sign indicates retardation or deceleration, meaning the object is slowing down. This change in velocity indicates non-uniform motion.

Problem 3:

Based on the provided position-time graph, calculate the velocity of the object between t=0 st = 0\text{ s} and t=4 st = 4\text{ s}.

Position-time graph showing a line passing through (0,0) and (4,8).

Solution:

From the graph, at t1=0 st_1 = 0\text{ s}, the position s1=0 ms_1 = 0\text{ m}. At t2=4 st_2 = 4\text{ s}, the position s2=8 ms_2 = 8\text{ m}. Using the formula for velocity: v=s2−s1t2−t1v = \frac{s_2 - s_1}{t_2 - t_1} v=8−04−0v = \frac{8 - 0}{4 - 0} v=2 m/sv = 2\text{ m/s} The velocity of the object is 2 m/s2\text{ m/s}.

Explanation:

Velocity is determined by the slope of the position-time graph. By picking two points on the line and dividing the change in position by the change in time, we find the constant speed.

Problem 4:

A car moves with a constant velocity of 10 m/s10\text{ m/s} for 5 s5\text{ s}. Draw the velocity-time graph and calculate the displacement during this interval.

Velocity-time graph showing a horizontal line at 10 m/s from time 0 to 5 seconds.

Solution:

The velocity-time graph is a horizontal line at v=10v = 10. Displacement ss is the area under the v−tv-t graph: Area=length×breadth\text{Area} = \text{length} \times \text{breadth} s=v×ts = v \times t s=10 m/s×5 ss = 10\text{ m/s} \times 5\text{ s} s=50 ms = 50\text{ m} The displacement of the car is 50 m50\text{ m}.

Explanation:

In a velocity-time graph for uniform motion, the displacement is the product of the constant velocity and the time duration, which corresponds geometrically to the area of the rectangle formed under the graph.

Plot and interpret position-time and velocity-time graphs for uniform motion Class 9 Notes &…