In this article, we will learn how to find acceleration in velocity time graph, using some examples, and solve some problems.

**The acceleration is the difference in the velocity varied with time; hence from the velocity-time graph, we can find the acceleration by measuring the slope of the graph.**

**Velocity Time Graph for Positive Acceleration**

Let us see, how to find the acceleration from the velocity-time graph. The following is a velocity v/s time graph.

The x-axis depicts the time in seconds and the velocity of the object at different time duration is given on the y-axis. **The slope of the graph is given by m=Δy/Δt. Here, the slope of the velocity-time graph gives the acceleration of the object.**

**a = m = ΔV/ΔT = v _{2}-v_{1}/t_{2}-t_{1}**

**From the above graph, the acceleration will be positive if V _{2}>V_{1} that is if the velocity of the object increases with time. The same will be negative if V_{2}<V_{1}, that is if the velocity of the object decreases with time.** This is the case when the object is decelerating. So is in the case, even when the object is traveling opposite from the direction of its motion.

**Read more on How To Find Acceleration With A Constant Velocity: Facts And Problem Examples.**

**Problem 1:** Consider a round shaped object at rest on the top of the hill. A force is applied to the object to displace it from its position. On application of force, the object accelerates downward to the bottom of the hill. The velocity of the object increases to 4m/s after covering a distance of 16 meters. Plot the graph for the same and then calculate the acceleration of the object considering the initial velocity of the object 2m/s at a certain time.

**Solution:** The variation in the velocity of the object is given as.

V_{1}=2m/s

V_{2}=4m/s

The velocity equal to 4m/s was seen after the object had traveled the distance of 16 meters. Therefore the time taken for the displacement of 16m and for the object to accelerate is

2m/s=16m/t

t=16m/2m/s=8s

Hence, the velocity of the object at time t= 8 seconds was 4m/s.Now we can plot a graph for the same as below

From the graph we have, velocity v_{1}=2m/s at t_{1}=4 sec and velocity v_{2}=4m/s at t_{1}=8 sec.

Therefore the acceleration of the object between the time interval 4sec to 8 sec is

a = v_{2}-v_{1}/t_{2}-t_{1} = 4-2/8-4 = 2/4 = 1/2 = 0.5m/s^{2}

The acceleration of the object is found to be 0.5m/s^{2}.

**Velocity-Time Graph for Zero Acceleration**

**The graph given below shows that the velocity of the object does not change with time and remains constant. This implies that there was no acceleration of the object between these time intervals.**

The above graph shows that the velocity of the object remains the same all the time hence we get a straight line in the velocity v/s time graph. It clearly indicates that the velocity-time graph doesn’t give slope in this case. Because there is no slope of the graph the acceleration which is equal to the slope is zero.

This means that the displacement of the object is the same for different time intervals hence the velocity is constant.

**Problem 2:**The velocity of the object moving on a plane surface was found to be 0.5 m/s. 5 minutes later it was found by another observer that the velocity was 0.5m/s. Then what is the acceleration of the object based on the observation?

**Solution:** V_{1}=0.5m/s; V_{2}=0.5m/s, Time interval t=5 minutes=300 seconds.

a=v_{2}-v_{1}/t_{2}-t_{1}= 0.5-0.5/300 =0

Since there was no variations in the velocity of the object was seen, the acceleration of the object is zero.

**Read more on acceleration.**

**Velocity-Time Graph for Negative Acceleration**

If the object decelerates with time, then the slope of the velocity-time graph will be negative. This is illustrated in the below velocity-time graph.

**Since the difference between the final and the initial point into consideration on the y-axis is negative, the slope of the graph which is the acceleration of the object will be negative.**

**Problem 3:** Consider an object decelerating with time as shown in the below graph.

Calculate the acceleration of the object from path A to B.

**Solution:** The velocity of the object at point A at time t_{1}=2seconds is v_{1}=10m/s and at time t_{2}=5 seconds is v_{2}=4m/s. Therefore the acceleration of the object is

a = v_{2}-v_{1}/t_{2}-t_{1} = 4-10/5-2= -6/3= -2m/s^{2}

Since the velocity of the object decreases with time, the acceleration of the object is negative and is found to be -2m/s^{2}.

**Read more on Constant Negative Acceleration Graph: What,How,Examples.**

**Negative Velocity Time Graph for Negative Acceleration**

When the object is moving away from its point of destination, in the negative axis, the displacement of the object is taken as negative on the negative y-axis. If the position of the object diverts away from its direction of motion then the displacement of the object is considered to be in a negative direction.

The above is the velocity-time graph for negative acceleration. **It is seen that the velocity is declining with time, the slope of the graph is found to be negative, and therefore the acceleration is negative.**

**Negative Velocity Time Graph for Positive Acceleration**

The following is a graph of negative velocity v/s time which gives the positive acceleration.

**As the decelerating object once starts accelerating back due to some external forces then the acceleration which is equal to the slope of the velocity v/s time graph is positive because the velocity of the object keeps on rising with time.**

**Read more on Negative Velocity And Zero Acceleration: How, When, Example And Problems.**

**Frequently Asked Questions**

**Q1. From the below graph calculate the acceleration of the object from point O to A, from A to B, and from B to C; and then calculate the average acceleration of the object from O to C.**

**Solution:** From O to A, v_{1}=0 at t_{1}=0; v_{2}=8m/s at t_{2}=4s

Therefore the acceleration of the object from point O to A is

a= v_{2}-v_{1}/t_{2}-t_{1}=8-0/4-0=8/4=2m/s^{2}

From A to B, v_{1}=8m/s at t_{1}=4s; v_{2}=5m/s at t_{2}=8s

Therefore the acceleration of the object from point A to B is

a=v_{2}-v_{1}/t_{2}-t_{1}=5-8/8-4=-3/4=-0.75m/s^{2}

From B to C, v_{1}=5m/s at t_{1}=8s; v_{2}=5m/s at t_{2}=12s

Therefore the acceleration of the object from point B to C is

a=v_{2}-v_{1}/t_{2}-t_{1}=5-5/12-8=0/4=0

The average acceleration of the graph from O to C is

A_{avg}= a_{oa}+a_{ab}+a_{bc}/3

=2-0.75+0/3=1.25/3=0.42m/s^{2}

Hence the average acceleration of the object from O to A is 0.42 m/s^{2}.

**Why acceleration is a vector quantity?**

The acceleration has magnitude and direction.

**The direction of the acceleration is the same as the direction of the velocity after a change; hence it is a vector quantity.**