
Collegedunia Team Content Curator
Content Curator | Updated On - Mar 17, 2025
V-T graphs in physics are a graphical representation of various types of motions ; it gives a quick description of the motion undertaken by the body/object of study. It is easier to study, formulate and therefore, is used to give quicker and simpler information about the state of a body. Based on the relative idea of rest and motion , the V-T graph deals with the velocity-time graph and provides details regarding the motion induced in the body in accordance with velocity, corresponding to time.
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Table of Content |
Key Terms: Velocity-Time Graph, Tyres, Equations of motion, Special Cases, Acceleration, Velocity
V-T Graph
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A vt graph is a graphical representation of different types of motions; the body/object under study exhibits motions by falling under a VT (velocity-time) frame. It means that the graph will discuss the motion of the body by seeing the change in its velocity with respect to time.
The y-axis represents the velocity of the object, x-axis represents the time and the slope of the graph is nothing but acceleration .
While there are three main equations of motion to represent, some other important parts remain too. Velocity-time graph represents the first equation of motion, v = u + at.
The three equations of motion are:
v = u + at.
v² = u² + 2as. (represented by the position-time graph)
s = ut + ½at² (represented by velocity-position graph)
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Various Cases of V-T Graph
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In this section we will check all the possible cases and graphical representations that we can draw with the help of a velocity-time graph.
When the object is at rest
When the object is at rest, it means there are two possibilities:
- Velocity, v = 0, or,
- Velocity, v = constant with respect to time.
Thus, the velocity-time graph will be like this:

This showcases that the object remains at rest and has no external force applied to change its state of motion.
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When Object is in Motion
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Once the object has been set in motion by some external force, there are mainly three cases based on the three equations of motions that the velocity-time graph represents:
When Object Moves with Uniform Velocity
When the object moves with uniform velocity, it means that the acceleration of the body is zero. Thus, it object moves with a constant velocity throughout with slope, i.e. acceleration, a = 0.
So, from
v = u + at
v = u, means initial velocity is equal to final velocity at all times.
The velocity axis forms a straight line to coincide with the time axis.

When Object moves with Uniform Acceleration
Uniform change in acceleration of the body means that the object changes its velocity at regular intervals of time. When object moves with uniform acceleration, there two further cases to study:
- When the object has positive (or upward) slope uniformly
When the object moves with positive or upward slope constantly, it showcases positive uniform acceleration. Means, the object accelerates at regular intervals in the direction of motion.

- When the object has negative (or downward) slope uniformly
When the object moves with negative or downward slope constantly, it showcases negative uniform acceleration. Means, the object de-accelerates at regular intervals opposite to the direction of motion.
- When object moves with changing acceleration
When acceleration of the body keeps changing, there are two more cases possible:
- When the object moves with positive (or increasing) acceleration
When the body moves with positive or increasing acceleration, it is generally the case when it moves forward. For example, a car driver peddling on the accelerator of the car to make it move ahead.

The magnitude of the slope continues to rise as acceleration increases . Also, initial velocity is often not zero in this case. Thus, by
v = u + at
u = 0
So, v = at making acceleration a function of time.
- When the object moves with negative (or decreasing) acceleration.
When the body moves with a negative acceleration, it is often termed as ‘retardation,’ means decreasing acceleration. It is generally the case when it starts to come to a halt.
For example, a car driver applying brakes for the car to either slowly come to stop or slow down the speed.

A changing acceleration may combine both accelerating and retarding forces to induce a change in the motion of the body by causing its velocity to fluctuate at irregular intervals of time. Thus, when a body moves with non-uniform acceleration, it is said to be in non-uniform motion.
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Special Cases
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Unlike other graphs, this one has its coordinates in the other quadrant too. So, what kind of graph is it?
It is a decelerating graph with constant acceleration. As the graph scales lower and touches 0 at the time axis, it represents the object has been de-accelerating and has finally come to stop. For example, a car driver applying brakes on the car and it stops.
So, v = 0
But, when the object further moves into the other quadrant, going beyond the time axis, it showcases that the object has begun moving again with constant acceleration, in the backward direction.
For example, a car driver reversing his car.
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Representation of First Equation of Motion
As we know velocity-time graph represents the first equation of motion, let’s derive it both algebraically and graphically:
Algebraic method:
As, acceleration = change in velocity/time
a = Δv/t
a = (v-u)/t
Rearranging gives us,
v = u + at.
Graphical method:
Consider a velocity-time graph like this:

- Velocity changes from A-B with respect to time, t (uniformly)
- Consider BC, v and OC, t
- Draw a perpendicular from B to OE and a parallel line is drawn from A to D. Thus, A perpendicular is drawn from B to OC.
BC = BD + DC
Hence,
v = BD + DC
and
v = BD + OA (since DC = OA)
Therefore,
v = BD + u (since OA = u) (Eq 1)
Now,
Let a = slope of line AB
[ { V(final) - V (initial) } ]/time = acceleration, a (this is the slope formula)]
a = BD/AD
Since
AD = AC = t,
BD = at (Eq 2)
Equation 1 + equation 2, we get:
v = u + at
Also Read: Difference Between Acceleration and Velocity
Things To Remember
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Here are some of the important things to remember:-
- The velocity-time graph is a graphical representation of the motion exhibited by a body with respect to the change in velocity , acceleration of slope by corresponding to time.
- When the object is in constant motion, the velocity-time graph is a straight line with no acceleration.
- When the object does have acceleration, if it is constant, it either goes to positive or negative slope depending upon acceleration and deceleration exhibited respectively.
- When the object has acceleration but it is non- uniform , it is either positive or negative depending upon the slope produced upon acceleration or deceleration respectively.
- To find the slope , perform { V(final) - V (initial) }/time, which will give acceleration.
Also Read: Derivation of Equation of Motion
Sample Questions
Ques. Calculate the maximum acceleration from the following velocity-time graph presented in m/s 2. ( 4 Marks)
Ans. Divide the graph into 4 parts and find acceleration accordingly { we know, acceleration, a = change in velocity/ change in time = v-u/t2-t1}.
For Part 1, initial velocity, u = 0 m/s and final velocity, v = 20 m/s
Time, t1 = 0 and t2 = 20 sec
So, a = 20-0/20-0
= 20/20
= 1 m/s 2
For Part 2, the velocity is constant throughout. Therefore, acceleration is zero.
For Part 3, initial velocity, u = 20 m/s and final velocity, v = 60 m/s
Time, t1 = 30 sec and t2 = 40 sec
So, a = 60-20/40-30
= 40/10
= 4 m/s 2
For Part 4, the body begins to decelerate after reaching its maximum velocity.
Initial velocity, u = 60 m/s and final velocity, v = 0 m/s
Time, t1 = 60 sec and t2 = 70 sec
So, a = 0-60/70-60
= -60/10
= -6 m/s 2
Hence, 4 m/s 2 in part 3 has the highest acceleration.
Ques. The displacement of a body = x = (t- 2) 2 where x is distance in metres and time is t is seconds. Find the distance covered by the particle in the first 4 secs. (4 Marks)
Ans. Since, we know that,
So, by differentiating, we get
v = dx/ dt = d/dt ( t-2) 2 = 2(t – 2) m/s
Now,
Acceleration, a = dv/dt = d [2(t-2)]
= 2[1-0]
= 2 m/s 2
Therefore, the plotted graph will be based on the special case graph because at time t = 0, the speed is 4 m/s while at t = 2, the body stops and comes backward. That is why, at t=4, the speed is 4 m/s in the opposite direction.
= area of triangle OAC + area of triangle ABD
= ½ x 4 x 2 + ½ x 2 x 4
= 8 m.
Ques. Calculate the distance travelled, represented by the red-line of the velocity-time graph. (4 Marks)
Ans. Distance travelled can be calculated by calculating the area of the region covered by the line in the following way:
- Calculate area of triangle
= ½ x base x height
= ½ x 4 x 8
= 16 m
- Calculate area of rectangle
= length x breadth {∴ length = 10-4 = 6}
= 6 x 8
= 48 m
So, total distance travelled is, area of triangle + area of rectangle
= 16 + 48
= 64 m
Ques. Calculate the acceleration shown by the purple-line of the velocity-time graph. (3 Marks)
Ans. For finding the acceleration of the purple-line, calculate the change in velocity with respect to time.
Initial velocity, u = 0 m/s and final velocity, v = 10 m/s
Initial time, t1 = 0 sec and final time, t2 = 2 sec.
Change in velocity = v-u = 10-0 = 10 m/s and change in time = t 2-t 1 = 2-0 = 2 sec
Acceleration, a = change in velocity / change in time
a = 10/2
= 5 m/s 2
Ques. Draw a V-t graph for an object that has been thrown upwards in projectile motion, with velocity v and comes back down in time t. (2 Marks)
Ans. As the object goes upwards, its velocity keeps decreasing. After reaching its maximum height, the velocity becomes zero and then it starts falling back. As it returns, its velocity goes on increasing.

That is why the special case graph is made. As you can see, the body first comes to rest after decelerating constantly (due to gravity) and then hits bottom at zero. Later on, it begins to bounce back that causes the graph to go beyond the time axis. It represents the motion of an object in the opposite direction.
Ques. Calculate the acceleration of the body as it changes its velocity uniformly from 20 m/s to 40 m/s. (2 Marks)
Ans. As uniform change in velocity means uniform acceleration:
Acceleration, a = change in velocity/change in time
= v 2-v 1/t 2-t 1
v 1 = 20 m/s and v 2= 40 m/s
t 1 = 2 sec and t 2 = 4 sec
So, a = 40-20/4-2
= 20/2
= 10 m/s 2
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