In this article, we are going to discuss some negative velocity examples and solve related problems.

**Some examples of negative velocity are as follow:-**

**Object falling down**

An object accelerating down from greater heights to down the surface decreases its position while accelerating. Since the position decreases with time, the velocity of the object is negative.

**The energy of the object is always conserved; it only changes from one form of energy to another. While the object accelerates upward its kinetic energy utilized for its drift is converted into potential energy and once all the kinetic energy is converted into the potential energy, the object starts accelerating down to zero potential surfaces.**

Since the energy is conserved in the form of the potential and kinetic energy of the object falling from the heights, its initial energy produced will be equal to the final energy.

[latex]KE_1+PE_1=KE_2+PE_2[/latex]

[latex]\frac{1}{2}m{u_1}^2+mgh_1=\frac{1}{2}m{v_2}^2+mgh_2[/latex]

[latex]\frac{1}{2}{u_1}^2+gh_1=\frac{1}{2}{v_2}^2+gh_2[/latex]

[latex]{u_1}^2+2gh_1={v_2}^2+2gh_2[/latex]

[latex]{u_1}^2-{v_2}^2=2g(h_2-h_1) [/latex]

[latex]{v_2}^2={u_1}^2-2g(h_2-h_1) [/latex]

[latex]{v_2}=\sqrt{{u_1}^2-2g(h_2-h_1)} [/latex]

In short, kinetic energy is equal to the potential energy of the object

[latex]\frac{1}{2}mv^2=mgh[/latex]

[latex]v^2=2gh[/latex]

[latex]v=\sqrt{2gh}[/latex]

**This implies that the velocity of the object during its flight in the air depends upon the height at which it is above the surface of the ground.**

As the height of the falling object decreases, the velocity of the object will decrease by the square root of its height. Hence the velocity of an object is negative in this case.

**Slipping of feet while trying to climb up the slider**

You must have noticed that due to lack of frictional force on the slider, we repeatedly slide backward while climbing up the slider from the slide. This **reduces the velocity of a person while climbing and often sliding back to the ground**. Hence, this is also an example of a negative velocity.

**Object slowing down**

**The fast accelerating object suddenly lowers its velocity then we have an exponential decrease in the velocity of the object.**

For such objects, the acceleration will also be negative.

**An object moving in a reverse direction with high speed**

**While moving in a reverse direction from the original point then the position of the object with respect to the origin will decrease, hence the velocity of the object which is a change in position with time will be negative.**

When the object takes sharp reverse acceleration, then the acceleration is positive but the velocity of the object is negative.

**Kingfisher diving in a river for fish**

A bird, kingfisher diving into the river water to catch the fish for its food moves with negative velocity as it comes near the surface of the water from the height at which it was flying.

**Light traversing through the medium**

As light travels from one medium to another, the speed of light changes. **On entering the denser medium the speed of light rays decreases, hence is also an example of negative velocity**, because the distance elapses by the light in the air in one second is more as compared to its velocity in the denser medium, as it travels slower comparably.

**Read more on the speed of light.**

**Spring**

Spring always tries to regain its original size upon stretching. If a heavy mass is attached to one end of the spring comparable to its spring constant and keeping the other end fixed, upon stretching the spring by pulling a mass on a horizontal slide, it will build potential energy which will be converted into kinetic energy once it is released.

**The mass will travel in a reverse direction due to equal and opposite reactions and will displace at a position towards the fixed point and remain stable. **This is an instantaneous process. Since the position decreases with time the velocity is negative.

**Flow of current**

**The direction of a flow of current is in the direction opposite to the motion of electron charges**; this also can be accounted for a negative velocity of electrons compared to the direction of the current.

**Solved Problems**

**Problem 1:** A car travels a distance of 7km in 7min and then slows down on reaching the district road and travels 3km in 9minutes. Calculate the change in velocity.

**Solution:** A car displaces 7 km from the original point, x_{1}=7km, time taken t_{1}=7min;

Hence the speed of the car is

[latex]v_1=\frac{x_1}{t_1}=\frac{7\times 60}{7}=60km/h[/latex]

A car travels 3 km in 9 minutes, hence the speed of the car is

[latex]v_2=\frac{x_2}{t_2}=\frac{3\times 60}{9}=20km/h[/latex]

Hence change in velocity is

[latex]V=V_2-V_1=20-60=-40km/h[/latex]

The change in the velocity of the car becomes negatives once it slows down.

**Problem 2:** A weight of mass 1kg is kept on a horizontal slide made up of glass. A weight is attached to a spring of length 1 meter whose other end is sealed in a wall. A mass is pulled to a distance of 50 cm away from its position on applying a force and relieved. After which the weight displaces 80 cm in a second and rests at a position 80 cm away from the wall.

Find the instantaneous velocity of a mass and the potential energy of spring on pulling the mass. Also, calculate the potential energy stored in the spring when it was displaced at a distance of 50 cm from its resting position. Given the spring constant k=1.5.

**Solution:**

Given: x_{1}=50cm, x_{2}=80cm, t_{1}=0, t_{2}=1sec;

[latex]Instantaneous\ Velocity=\frac{x_2-x_1}{t_2-t_1}=\frac{50-80}{1-0}=-30cm/s=-0.3m/s[/latex]

Velocity of a weight is -0.3m/s which are negative because the weight displaces in the reverse direction.

Problem 3: Find the average velocity of an object traveling with time. The distance elapsed by the object at different times was noted, the same is shown in the following table. Plot the graph for the same.

Displacement(km) | Time(min) |

10 | 10 |

8 | 20 |

6 | 30 |

4 | 40 |

**Solution:** The position-time graph for the above data is as below

Change in velocity with respect to time is equal to the slope of the graph.

[latex]v_1=\frac{x_2-x_1}{t_2-t_1}=\frac{8-10}{20-10}=\frac{-2}{10}=\frac{-2\times 1000}{10\times 60}m/s=-3.3m/s[/latex]

[latex]v_2=\frac{x_2-x_1}{t_2-t_1}=\frac{6-8}{30-20}=\frac{-2}{10}=\frac{-2\times 1000}{10\times 60}m/s=-3.3m/s[/latex]

[latex]v_3=\frac{x_2-x_1}{t_2-t_1}=\frac{4-6}{40-30}=\frac{-2}{10}=\frac{-2\times 1000}{10\times 60}m/s=-3.3m/s[/latex]

[latex]v_4=\frac{x_2-x_1}{t_2-t_1}=\frac{4-8}{40-20}=\frac{-4}{20}=\frac{-4\times 1000}{20\times 60}m/s=-3.3m/s[/latex]

It is seen that the velocity of the object remains constant but the position of the object decreases with increasing time, hence the velocity of the object is negative.

The potential energy of a spring is half time the spring constant and a square of the displacement.

[latex]U=\frac{1}{2}kx^2[/latex]

[latex]=\frac{1}{2}\times 1.5\times {0.5}^2[/latex]

[latex]=\frac{1}{2}\times 1.5\times 0.25[/latex]

[latex]=0.19\ Joules[/latex]

Hence, the energy stored in the string was 0.19 Joules.

**Problem 4:** A girl walks 100meters towards North in 1minute and returns back to the point from where she had started and walks 200meters towards South from there in 4minutes. Calculate her actual velocity and displacement to reach that point where now she is.

**Solution:** A girl walks 100meters initially towards North in 1minute hence her walking speed was

[latex]Speed=\frac{Distance}{Time}[/latex]

[latex]V_1=\frac{100}{60}=1.67m/s[/latex]

Then, a girl covers 200meters in 4minutes, therefore the velocity of a girl is

[latex]v_2=\frac{200}{4\times 60}=0.83m/s[/latex]

Therefore, the velocity of a girl in 5minutes is

[latex]V=V_2-V_1=0.83-1.67=-0.84m/s[/latex]

And actual displacement of a girl from original position is

[latex]x=x_2-x_1=200-100=100meters[/latex]

That is 100 meters towards the South from the original position.

**Problem 5:** What is the velocity of an object when it is at a height of 10 meters above the ground and how does the velocity of the object vary at what rate?

**Solution:** The velocity of the object during its flight in the air is proportional to the square root of its height from the ground and the acceleration due to gravity of the Earth given by the relation

[latex]v=\sqrt{2gh}[/latex]

When the object is at a height of 10 meters, it velocity will be equal to

[latex]v=\sqrt{2\times 9.8\times 10}=\sqrt{196}=14 m/s[/latex]

When h=9m

[latex]v=\sqrt{2\times 9.8\times 9}=\sqrt{176.4}=13.28 m/s[/latex]

When h=8m

[latex]v=\sqrt{2\times 9.8\times 8}=\sqrt{156.8}=12.52 m/s[/latex]

When h=7m

[latex]v=\sqrt{2\times 9.8\times 7}=\sqrt{137.2}=11.71 m/s[/latex]

When h=6m

[latex]v=\sqrt{2\times 9.8\times 6}=\sqrt{117.6}=10.84 m/s[/latex]

When h=5m

[latex]v=\sqrt{2\times 9.8\times 5}=\sqrt{98}=9.8 m/s[/latex]

When h=4m

[latex]v=\sqrt{2\times 9.8\times 4}=\sqrt{76}=8.7 m/s[/latex]

When h=3m

[latex]v=\sqrt{2\times 9.8\times 3}=\sqrt{58.8}=7.67 m/s[/latex]

When h=2m

[latex]v=\sqrt{2\times 9.8\times 2}=\sqrt{39.2}=6.26 m/s[/latex]

When h=1m

[latex]v=\sqrt{2\times 9.8\times 1}=\sqrt{19.6}=4.43 m/s[/latex]

Height (m) | Velocity(m/s) | Acceleration |

10 | 14 | -0.72 |

9 | 13.28 | -0.76 |

8 | 12.52 | -0.81 |

7 | 11.71 | -0.87 |

6 | 10.84 | -1.04 |

5 | 9.8 | -1.10 |

4 | 8.7 | -1.03 |

3 | 7.67 | -1.41 |

2 | 6.26 | -1.83 |

1 | 4.43 | -4.43 |

This indicates that, as the height decreases, the velocity of an object decreases, and hence acceleration also decreases. The acceleration of the object sharply decreases when it is near the ground surface.

**Problem 6:** A ray of light enters a bucket of water and passes through a glass slab placed in it. Find the velocity of the light rays in each medium. The refractive index of water is 1.3 and that of glass is 1.5.

**Solution:** RI of water is n=1.3,

[latex]n=\frac{c}{v}[/latex]

[latex]v=\frac{c}{n}=\frac{3\sqcap 10^8}{1.3}[/latex]

[latex]v=2.3\times 10^8\ m/s[/latex]

Now the velocity of light is [latex]2.3\times 10^8\ m/s[/latex]. Hence, the speed of light in glass is

[latex]n=\frac{v_1}{v_2}[/latex]

[latex]v_2=\frac{v_1}{n}=\frac{2.3\times 10^8}{1.5}[/latex]

[latex]v=1.5\times 10^8\ m/s[/latex]

Hence the velocity of light in glass is [latex]1.5\times 10^8\ m/s[/latex].

**Read more on Negative Velocity Graph: Different Graphs And Their Explanations.**

**Frequently Asked Questions**

**Why velocity is negative?**

Velocity is defined as a change in the position of an object with different time intervals.

**If the position of an object shifts with time, the difference in position with respect to its previous position will be negative and hence velocity will be negative compared to the preceding interval.**

**How phase velocity of light can be negative?**

Distance traveled by a single ray of light or single wave in one second of time is called phase velocity.

**Negative velocity**** comes into a scene while light travels from one medium to another because the speed of light decreases on entering the denser medium.**