In this article, we are going to discuss negative velocity along with graphs and solve problems to understand various facts of negative velocity.

**If the velocity of an object decreases with respect to time duration then the object is said to have a negative velocity that might be constant, varying, instantaneous, or relative to the direction of the velocity of some other object into consideration.**

**Constant Negative Velocity Graph**

If the slope of the position v/s time graph is negative and the distance decreases at a constant rate along with time, then it is said to be a constant negative velocity graph.

**If ‘m’ is a value equal to the slope of the graph, which is linear and remains the same on calculating the slope between any two points on the line then the velocity is constant.** Since the distance abates with time, the slope is negative and hence the velocity is negative.

**Problem 1:** Consider a pulley ties with two masses on both the ends of the rope of length 30meters, the mass on one end of the rope is pulled to raise another mass tied on another end of the rope at a constant rate. If 10meters of the rope was pulled in the first 12seconds and 20meters of ropes in the next 24seconds, then calculate the velocity of the mass tied on another end.

**Solution:** Since the change in length of the rope on one side in 12seconds is from say 0 to 10 meters and at the same time length of the rope has decreased to 30-10=20meters on another end of the rope.

Now the length of the rope is 20meters, so, after pulling 20meters of rope in 24seconds, the length of the rope on the other side is 20-20=0meters.

Hence, we have: x_{1}=30m, x_{2}=20m and x_{3}=0

Time t_{1}=0, t_{2}=12 seconds, t_{3}=12+24=36seconds

Therefore,

Slope_{1}=x_{2}-x_{1}/t_{2}-t_{1}=20-30/12-0=-10/12=-0.8m/s

Slope_{2}=x_{3}-x_{2}/t_{3}-t_{2}=0-20/36-12=-20/24=-0.8m/s

Slope_{3}=x_{3}-x_{1}/t_{3}-t_{1}=0-30/36-0=-30/36=-10/12=-0.8m/s

It is clear that

Slope_{1}=Slope_{2}=Slope_{3}=-0.8m/s

Hence the slope is linear and has constant negative velocity.

**Negative Uniform Velocity Graph**

**When the object covers equal distance in equal intervals of time then the object is said to have uniform velocity, and if the object traverses back in a uniform velocity then the object is moving with negative uniform velocity.**

As the object displaces an equal distance in an equal interval of time, it implies that the velocity of the object is constant and hence there is no acceleration of the object.

**Problem 2:** Supposed, a narrow road runs from a village from point A to B such that only one car can travel on the road at a time. The length of the narrow road is 1km long. A carA travels a distance of 300meters from a narrow road from point A and encounters carB, hence starts reversing back at a constant speed and covers 3meters per second. Plot a graph and find the velocity at 3 different points.

**Solution:** The displacement of the car is 3meters per second and the distance from pointA decreases at the rate of 3m/s. A CarA had covered a distance of 300meters and will traverse back 300meters at a rate of 3m/s.

Position(x meters) | Time(t sec) |

300 | 0 |

297 | 1 |

294 | 2 |

291 | 3 |

288 | 4 |

285 | 5 |

**Table showing position of an object varied with time**

We plot a graph for displacement v/s time,

Slope_{1}=x_{2}-x_{1}/t_{2}-t_{1}=294-297/2-1=-3/1=-3m/s

Slope_{2}=x_{3}-x_{2}/t_{3}-t_{2}=291-294/3-2=-3/1=-3m/s

Slope_{3}=x_{3}-x_{1}/t_{3}-t_{1}=288-291/4-3=-3/1=-3m/s

Slope_{1}= Slope_{2}= Slope_{3}=-3 m/s

The slope is constant and negative, and hence the velocity of the car taking a reverse on a narrow road is -3m/s.

**Negative Relative Velocity Graph**

**Velocity is a vector quantity and relative velocity is a vector difference of velocities of two bodies. That is if the velocity of object A is V _{a} and that of object B is moving with velocity V_{b}, then the relative velocity of both the objects with respect to each other is V_{ab}=V_{a}-V_{b}.**

On finding the slope of position v/s time plot, we can calculate the relative velocity of the object with respect to each other. If both the objects are decelerating then the slope of the graph will be negative.

**Problem 3:** A car traveling with a velocity of 60km/hr crosses a woman walking on a street with a speed of 2m/s in the same direction. What is their relative speed?

**Solution:** V_{1}=60 km/h=60*1000/60*60=16.67m/s

V_{2}=2m/s

Hence, the relative velocity of the car with respect to a lady is

V=V_{1}-V_{2}=60-2=58m/s

The relative velocity of the car will be 58m/s and that of the woman will be –58m/s as the speed of a car is faster than the woman.

**Negative Velocity Positive Acceleration Graph**

**If the object accelerates back from its original position along with time by changing the velocity of the object then we have negative velocity but the acceleration of the object is positive.**

The object accelerating backward changes its velocity frequently, that is the rate of displacement of the object in a time interval is not constant and hence the graph shows different slopes on plotting the position of the object at different times.

**If the velocity of the object decreases at an exponential rate then we get the positive acceleration from the negative velocity of the object.** Let us illustrate this with the problem below.

**Problem 4:** Consider the same situation given in problem No.2 and the same car is accelerating backward decreasing its speed. Suppose the covers first 200 meters with a speed of 40 km/h and next 50 meters with a speed of 15 km/h and remaining distance in 10 km/h speed. Then calculate the velocity and acceleration of the car at different points.

**Solution:** A car initially was at say point X_{1}= 300 and travels 200 meters to reach point X_{2}= 100 with velocity V_{1}=40 km/h. From where the velocity of the car changes to V_{2}=15 km/h and travels next 50 meters and velocity slightly drop to 10 km/h and comes at the original position on traveling 50 meters on the same speed.

The time taken to elapse 200 meters with the speed of 40 km/h is

t_{1}=Distance/Speed=200*60*60/40*1000=18 seconds

Time taken to cover 50 meters with a speed of 15 km/h is

t_{2}=Distance/Speed=50*60*60/15*1000=12 seconds

And a time taken to cover a distance of 50 meters with a speed of 10 km/h is

t_{3}=Distance/Speed=50* 60*60/10*1000=18 seconds

Therefore, T_{1}=18 seconds, T_{2}=18+12=30 seconds, T_{3}=30+18=48 seconds

Since the acceleration is define as the change in the velocity of a car in different time intervals, hence,

a_{1}=v_{2}-v_{1}/Time Interval=40-15/18=25*1000/18*60* 60=0.38 m/s^{2}

a_{2}=v_{2}-v_{1}/Time Interval=15-10/12=5*1000/12*60*60=0.12 m/s^{2}

a_{3}=v_{2}-v_{1}/Time Interval=10-0/18=10*1000/18*60*60=0.15 m/s^{2}

This gives the positive acceleration. At the steeper slope of graph, the acceleration is of a car is higher and at gentle slope the acceleration is smaller.

**Instantaneous Velocity Negative Graph**

**The object is said to have an instantaneous velocity when it displaces from its place drastically. If the displacement is in a reverse direction then it is said to have instantaneous negative velocity.**

**Here, the displacement of the object is seen suddenly in a short span of time and hence the instantaneous velocity is high.**

This is seen in the case of spring tied with a mass at one end having its own potential and another end of the spring is held fixed on the rigid wall. When the mass is pulled away from the spring, the spring potential energy is built and the spring force pulls back the mass towards it to regain its original size by converting this potential energy into kinetic energy.

If the mass is heavy, then after releasing the mass attached to the spring, the spring force displaces the mass towards the rigid wall. The mass will resist the spring force and makes its place there. Hence, the position of the mass varied, and the distance separating it from the rigid wall decreased.

**Since the displacement decreases, we have negative velocity in the picture.**

**Problem 5:** If the mass of 2kg is attached to a string of length 1.5 meters at one end and another end is fixed at a rigid wall. On pulling the 50 cms away from its position linearly and released, the mass displaces towards the wall and remains stable at 80 cm away from the wall. Find the instantaneous velocity of the mass if the mass came to its resting position in 1 second.

**Solution:** A mass is displaced 150 – 80 =70cms =0.7m from its original position and has covered a distance of 70+50 =120cms =1.2m on releasing the spring.

Instantaneous velocity =Displacement/Time taken=1.2 m/1 second=1.2 m/s

Hence, the mass is displaced with a velocity of 1.2 m/s.

**Negative Velocity v/s Time Graph Displacement**

The velocity is a ratio displacement in a specific time duration, given by the relation

Velocity=Displacement/Time

Hence, **the displacement of an object in time ‘t’ moving with velocity ‘v’ is**

**x=vt**

**The displacement of the object at a point or at a certain time can be calculated by multiplying it with the velocity of the object at that time.**

**Problem 6:** Based on the following graph calculate the distance between points A & B.

**Solution:** Distance of Point A from the origin is

x_{1}=v_{1}t_{1}=10* 50=500m

The distance between point B and the origin is

x_{2}=v_{2}t_{2}=20*30=600m

Hence, the distance between point A & point B is

x=x_{2}-x_{1}=600-500=100m

Therefore, point B is 100 meters away from point A.

**Read More on Negative Refraction.**

**Negative Slope Velocity Time Graph**

**The slope of the velocity-time graph will be negative only when the velocity of an object traveling decreases along with time.**

If the slope of the velocity-time graph is negative, it means the acceleration is negative.

**Problem 7:** Find the average acceleration of a swing whose velocity is decreased with time as shown in the table.

Velocity(m/s) | Time(seconds) |

8 | 1 |

6 | 5 |

4 | 10 |

2 | 15 |

**Solution:** Let’s calculate the acceleration at different time intervals

a1=v_{2}-v_{1}/t_{2}-t_{1}=4-8/10-1=-4/9=0.44 m/s^{2}

a2=v_{2}-v_{1}/t_{2}-t_{1}=2-6/15-5=-4/10=0.40 m/s^{2}

a3=v_{2}-v_{1}/t_{2}-t_{1}=6-8/5-1=-2/4=0.5 m/s^{2}

a4=v_{2}-v_{1}/t_{2}-t_{1}=4-6/10-5=-2/5=0.40 m/s^{2}

Hence, the average acceleration is

aˉ=a_{1}+a_{2}+a_{3}+a_{4}/4

aˉ=0.44+0.4+0.5+0.4/4=0.435 m/s^{2}

**Negative Displacement Velocity Time Graph**

**If the object is taking a reverse turn from its original position with decreasing velocity along with time then we get negative displacement velocity on plotting the same on a graph.** The same is demonstrated in the graph below.

Since the velocity of the object is equal to the displacement of the object it makes in time duration, the displacement can be calculated as a product of the velocity of the object into time.

**Problem 8:** Consider the above velocity-time graph; the object is decelerating with time. Based on the above graph calculate the displacement of the object at a time= 5 seconds.

**Solution:** AT time t=5seconds, v=-20 m/s.

Velocity =Displacement/Time

x=vt

x=-20*5=-100m

Hence the displacement of an object is -100 meters from origin.

**How to Calculate Distance from Negative Velocity Time Graph?**

**Since the velocity of the object is determined by the distance it covers in a specific time, the displacement of the object is a product of its velocity into time.**

Velocity =Displacement/Time

x=vt

**The displacement from the velocity-time graph is the area covered by the curves in the graph.** For negative velocity the displacement will also be calculated and found to be negative, henceforth the displacement of the object with respect to its original position can be known.

Lets us see, how to calculate the displacement from the negative velocity-time graph through an example.

**Problem 9:** Consider the following velocity-time plot of an object, based on it calculate the displacement of an object and its position from its original position.

**Solution:** The area of the triangle in first quadrant is

Area of triangle =1/2 bh

x_{1}=1/2*10* 7=35m

The area of the triangle in fourth quadrant is

x_{2}=1/2*12* (-8)=-48m

Hence, the total displacement of an object is

x=x_{1}+x_{2}=35-48=-13m

This implies that the object has been displaced 13 meters further from the original position.

This is how the displacement is calculated from the velocity-time graph.

**Read more on How To Find Final Velocity Without Acceleration: Facts, Problems, Examples.**

**Frequently Asked Questions**

**How will you plot a graph from an object accelerating at a speed of 2m/s with its initial velocity of 4m/s?**

**Given:** a=2m/s, u=4m/s

Velocity and time relation is given by the formula

v=u+at

At time t=0,

v=4+2*0=4m/s

At time t=1,

v=4+2* 1=4+2=6m/s

At time t=2,

v=4+2*2=4+4=8m/s

At time t=3,

v=4+2*3=4+6=10m/s

At time t=4,

v=4+2*4=4+8=12m/s

At time t=5,

v=4+2* 5=4+10=14m/s

Time(sec) | Velocity(m/s) |

0 | 4 |

1 | 6 |

2 | 8 |

3 | 10 |

4 | 12 |

5 | 14 |

Plotting the graph of the same,

**How to find acceleration from a velocity–time graph?**

Acceleration is a rate of change of velocity at different time intervals.

**Hence the slope of a graph of velocity v/s time will give the acceleration of the body.**