Kinematics(displacement and Velocity)

[BlueOwl]

Homework Statement

A car starts from a point and returns to the same point after 160 seconds. During this time the car traveled at an average speed of 10 m/s. What is its displacement? What is its average velocity? Explain your answers using a motion graph (Position - time)

Homework Equations

The Attempt at a Solution

Ok so I am a little confused with this one since the car starts from the same point it returns to therefore i was assuming that the displacement would be zero and considering that average velocity is displacement over change of time zero divided by 160 again the average velocity would be zero no? please help I am really confused by the same points meaning the car is going from point A to Point A again not Point A to Point B therefore i am tottally lost maybe a hint to get me started thanks .

 

[alphysicist]

Hi BlueOwl,

Homework Statement

A car starts from a point and returns to the same point after 160 seconds. During this time the car traveled at an average speed of 10 m/s. What is its displacement? What is its average velocity? Explain your answers using a motion graph (Position - time)

Homework Equations

The Attempt at a Solution

Ok so I am a little confused with this one since the car starts from the same point it returns to therefore i was assuming that the displacement would be zero and considering that average velocity is displacement over change of time zero divided by 160 again the average velocity would be zero no?

Those answers sound right to me.

 

[nlightner]

Hi there,

Thanks for reaching out. I can see why this question might be a bit confusing, but let me try to explain it to you.

First, let's define displacement and velocity. Displacement is the change in an object's position from its starting point to its ending point. Velocity, on the other hand, is the rate at which an object changes its position. It is a vector quantity, meaning it has both magnitude (speed) and direction.

In this problem, the car starts at a point and returns to the same point after 160 seconds. This means that its starting and ending positions are the same, so its displacement is indeed zero. However, this does not mean that its average velocity is also zero.

To calculate average velocity, we need to use the formula v = Δx/Δt, where v is the average velocity, Δx is the change in position, and Δt is the change in time. In this case, since the car starts and ends at the same point, Δx is zero. However, the car still traveled a distance of 160 seconds, so Δt is 160 seconds. Plugging these values into the formula, we get:

v = 0/160 = 0 m/s

This means that the car's average velocity is indeed zero, which makes sense because it did not change its position.

Now, let's look at the motion graph (position-time) for this situation. The x-axis represents time and the y-axis represents position. Since the car starts and ends at the same point, the graph will be a straight horizontal line at the y-axis, as shown below:

Position - Time Graph:

-------------------
| |
| |
| |
| |
| |
|___|___|

0 160

As you can see, the car's position does not change over time, so the slope of this line is zero. This slope represents the car's velocity, so it confirms our calculation of zero average velocity.

I hope this helps clarify things for you. Let me know if you have any other questions.

Best,