Graphical Representation of Motion - Velocity - Time Graphs

Activity:

The train moves at a constant velocity of 60 km/h for 5 hours, which means the velocity remains the same throughout.

• To determine the distance covered between 2 and 4 hours, look at the area under the velocity-time graph between these two time points.
• The area under a velocity-time graph represents the distance travelled. In this case, it's a rectangle because the velocity is constant.
• The distance covered between 2 and 4 hours is equal to the area of the rectangle formed by the velocity of 60 km/h and the time interval from 2 to 4 hours.

The formula for distance is Distance = Velocity × Time.

In this case, Distance = 60 km/h × (4 - 2) hours = 60 × 2 = 120 km.

Since the train is moving at uniform velocity, there is no acceleration. Acceleration is zero because the speed doesn't change over time.

Velocity-time graph

To draw velocity-time graphs, we will use the three equations of motion.

Case 1: Velocity-time graphs with constant velocity (zero acceleration)

When the velocity is constant, the velocity-time graph, with the Y-axis denoting velocity and the X-axis denoting time, will be like:

The graph clearly shows that the velocity remains constant at c throughout the time interval. No particles of matter how much the time changes, the velocity will be c at every instant. In this case, we have taken the initial velocity to be positive. The graph will be different if the initial velocity is negative.

Example: If the acceleration of a particle is zero (0), and velocity is constantly, say, 5 m/s at t = 0, then it will remain constant throughout the time.

Case 2: Velocity-time graphs with constant acceleration

When the acceleration is constant (positive), and the initial velocity of the particle is zero, the velocity of the particle will increase linearly as predicted by the equation:

v = u + at

Since u = 0

v = at

As shown in the figure, the velocity of the particle will increase linearly with respect to time. The slope of the graph will give the magnitude of acceleration.

Example: If the acceleration of a particle is constant (k) and is positive, the initial velocity is zero, and then the velocity increases linearly. The slope of the velocity-time graph will give the acceleration.

Case 3: Velocity-time graphs with increasing acceleration

When the acceleration is increasing with time, the velocity-time graph will be a curve as predicted from the equation:
v = u + at
Since u = 0
v= at

Since acceleration is a function of time, the velocity-time graph will be a curve.

Note: Since the acceleration is continuously increasing with time, the magnitude of the slope will also continuously increase with time.

When an object moves along a straight line with uniform acceleration, it is possible to relate its velocity, acceleration during motion and the distance covered by it in a certain time interval by a set of equations known as the equations of motion. For convenience, a set of three such equations is given below:
v = u + at                          ......(1)
s = `"ut" + 1/2at^2`                  ......(2)
`2 as = "v"^2 – "u"^2`                   ......(3)

where u is the initial velocity of the object that moves with uniform acceleration a for time t, v is the final velocity, and s is the distance travelled by the object in time t. Eq. (1) describes the velocity-time relation and Eq. (2) represents the position-time relation.

Eq. (3), which represents the relation between the position and the velocity, can be obtained from Eqs. (1) and (2) by eliminating t.

These three equations can be derived by graphical method.

The table shows the velocity of the car at different times, increasing by 8 m/s every 5 seconds. This means the car is accelerating uniformly, and the velocity increases by equal amounts in equal time intervals.

The velocity change every 5 seconds is 8 m/s, so in 5 minutes (300 seconds), the total change in velocity would be:

• Number of 5-second intervals = `"300"/"5"`=60
• Total velocity change = 60 × 8 = 480 m/s.

The velocity-time graph is a straight line because the car is uniformly accelerating. In non-uniformly accelerated motion, the graph would be curved, showing varying acceleration over time.

To calculate the distance covered between the 10th and 20th seconds, we need to find the average velocity in that interval.

The average velocity is the mean of the velocities at 10 seconds (16 m/s) and 20 seconds (32 m/s), which is `"16+32"/"2"`=24 m/s.

The distance is then:
Distance = Average velocity × Time = 24 × 10 = 240

Check that, similar to the example of the train, the distance covered is given by the area of quadrangle ABCD.

The area can be split into two parts:

1. Rectangle AECD

Base = 10 seconds (from 10s to 20s)

Height = 16 m/s (velocity at 10 seconds)

Area = 10×16=160 m

2. Triangle ABE

Base = 10 seconds (time from 10s to 20s)

Height = 16 m/s (difference in velocity from 16 m/s to 32 m/s)

Area = 12×10×16 = 80 m

Total distance = Area of rectangle + Area of triangle = 160 m + 80 m = 240 m.

The total area of quadrangle ABCD gives the distance travelled, confirming the calculation.

Velocity-time graph