Acceleration-Time Graphs

Motion can be defined as a shift in an object’s position over time. An athlete running on a track, the hands of a clock, a moving train, the needle of a sewing machine, etc. are some examples of bodies in motion. The motion of an object can be mathematically defined in terms of distance, displacement, velocity, acceleration, and time. Depending on the nature of the movement, the motion of a body is classified into four types: linear motion, rotary motion, oscillatory motion, and reciprocating motion.

Linear motion: The term “linear motion of a body” refers to a movement along a straight path. Some examples of linear motion are a cyclist cycling on a straight road, a swimmer swimming in a straight line, and the motion of a bus on a straight road.

Rotary motion: The rotation of a body along a fixed axis is referred to as rotary motion. Some examples of rotary motion are the blades of a fan, the wheels of a moving car, and the motion of a ferry wheel.

Oscillatory motion: The to-and-fro motion of a body about its mean position is referred to as oscillatory motion. Some examples of oscillatory motion are a pendulum clock, a tuning fork, and the motion of a swing.

Reciprocating motion: A body moving back and forth or up and down repeatedly is known as a reciprocating motion. Some examples of reciprocating motion are a shaping machine, the needle of a sewing machine, and an electric doorbell.

### Acceleration-time graphs

An acceleration-time graph is used to depict the relationship between acceleration and time, where acceleration is an independent variable plotted on the Y-axis, and time is a dependent variable plotted on the X-axis. It is obtained when the data of acceleration and time obtained is plotted on a rectangular graph while studying the motion of a body. We can determine the change in velocity in a given interval of time with the help of an acceleration-time graph. By determining the area under the curve of the acceleration-time graph, we can find the change in velocity of an object in motion over a certain period.

The graphs shown above are acceleration-time graphs for various types of body motion.

1. When a body is moving with constant acceleration,
2. When a body is moving uniformly with increasing acceleration.
3. When a body is moving non-uniformly with variable acceleration.

#### Conclusions

The following are the points concluded from the acceleration-time graphs.

• The slope of the acceleration-time graph is △a/△t.
• The area under the acceleration-time graph represents the change in velocity.

Let,

△v be the change in velocity,

△a be the change in acceleration,

△t be the change in time.

Now, the area under the curve = △v

We know that the acceleration of a body is referred to as the rate of the change in velocity in a given period of time.

So,

△a = △v/△t

Now, by multiplying with △t on both sides, we get,

△v = △a × △t (which is area under the curve in acceleration-time graph)

• For an object moving with constant acceleration, i.e., acceleration is the same at all intervals of time. Hence, the slope of the acceleration-time graph is zero, and the graph will be parallel to the time axis.

#### Acceleration vs Retardation

Acceleration: Acceleration of a body is referred to as the rate of the change in velocity in a given period of time. It is a vector quantity and is measured in terms of m/s2.

Acceleration = Change in velocity/time

here, change in velocity is positive.

Let us assume that a body that is traveling at a uniform velocity increases its speed. The rate of increase in speed is referred to as the acceleration of a body. Let “u” be the initial velocity, “v” be the final velocity, and “t” be the time interval, then the acceleration of the body is given as,

Acceleration (a) = (v – u)/t

Example: An example of acceleration in everyday life is turning a car at a corner. The velocity of the car increases as the direction changes, which results in acceleration. The acceleration increases with increasing turn speed.

Retardation: Retardation of a body is referred to as the rate of the change in velocity in a given period of time. Retardation means negative acceleration.

Retardation = Change in velocity/time

here, change in velocity is negative.

Let us assume that a body that is traveling at a uniform velocity decreases its velocity. The rate of decrease in velocity is referred to as the retardation of a body. Let “u” be the initial velocity, “v” be the final velocity, and “t” be the time interval, then the retardation of the body is given as,

Retardation = (v – u)/t

Example: A real-life example of retardation is turning off a fan. When the switch of moving fan is switched off, the fan blades’ rate of rotation progressively slows down. This rate of reduction in the velocity of the fan is known as retardation.

### Sample Problems

Problem 1: From the acceleration vs time graph given below, determine the change in velocity.

Solution:

To find the change in velocity of the body we need to determine the area under the curve.

So, to find the change in the velocity of the object, we need to calculate the area of the triangle.

△v = area of triangle

= ½ × 25 × 6

△v = 75 m/s.

Hence, the change in velocity is 75 m/s.

Problem 2: From the acceleration vs time graph given below, find the initial velocity of a body if its final velocity is 55 m/sec.

Solution:

To find the change in velocity of the body we need to determine the area under the curve. From the graph, we can observe that it has a rectangle and a triangle. So, to find the change in the velocity of the object, we need to calculate the area of these figures.

△v = area of triangle + area of rectangle

= ½ × 8 × 6 + 2 × 8

△v = 24 + 16 = 42 m/s

We know that,

△v = final velocity – initial velocity

42 = 55 – vi

vi = 55 – 42 = 13 m/s

Hence, the initial velocity is 13 m/s.

Problem 3: From the acceleration vs time graph given below, determine the velocity at t = 6 sec, if v(0) = 0.

Solution:

Acceleration is given by;

a = dv/dt

⇒ dv = (a)dt

By integrating on both sides, we get

∫dv = ∫(a) dt

⇒ v = ∫(1.5) dt

⇒ v(t) = 1.5t + c, where c is a constant

⇒ v(0) = 0

⇒ c = 0

Now, v(t) = 1.5t

v(6) = 1.5 × 6 = 9 m/s

Hence, the velocity at t = 6 sec is 9 m/s.

Problem 4: What does the area under the acceleration-time graph indicate?

Solution:

The area under the acceleration-time graph represents the change in velocity.

Let △v be the change in velocity, △a be the change in acceleration, and △t be the change in time.

Now, the area under the curve = △v

We know that the acceleration of a body is referred to as the ratio of the change in velocity in a given period of time.

So,

△a = △v/△t

Now, by multiplying with △t on both sides, we get,

△v = △a × △t

So, the area under the curve is obtained by multiplying the change in acceleration and change in time.

Problem 5: What does a jerk in the acceleration time graph mean?

Solution:

A jerk is the sudden change in the acceleration of the moving body, and the slope of the a-t graph represents jerk.

The slope of the acceleration-time graph = jerk = △a/△t

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