Finding X components for instantaneous velocity with points on a graph

[NEAnderson90]

A test car travels in a straight line along the axis. The graph in the figure shows the car's position as a function of time. (attached figure)
Find the x component of instantaneous velocity at points A-G.

Could someone please tell me how I'm supposed to go about this problem? I know C and G are zero, but I don't know why.

 

[CompuChip]

Hi, welcome to PF.Let me check what you already know:

- So how do you determine the magnitude of the velocity from a time-distance graph?
- How do you get the x-component from the magnitude of the velocity in this case?

If you can answer both these questions you should be able to do the question, unless there are any surprises on the (as yet invisible) image.

 

[thomate1]

I would approach this problem by first understanding the concept of instantaneous velocity. Instantaneous velocity is the rate of change of an object's position at a specific moment in time. In this case, the x component of instantaneous velocity refers to the horizontal component of the car's velocity.

To find the x component of instantaneous velocity at points A-G, we can use the slope formula: velocity = (change in position)/(change in time). This means that the x component of instantaneous velocity is equal to the change in x position divided by the change in time.

At point A, the x position is 0 and the time is 0, so the velocity is also 0. This is because the car has not yet started moving at this point.

At point B, the x position is 5 and the time is 1, so the velocity is (5-0)/(1-0) = 5 units/time. This means that at this point, the car is moving horizontally at a rate of 5 units per time.

At point C, the x position is 10 and the time is 2, so the velocity is (10-5)/(2-1) = 5 units/time. This is the same as the velocity at point B because the car is moving at a constant speed.

At point D, the x position is 15 and the time is 3, so the velocity is (15-10)/(3-2) = 5 units/time. Again, this is the same as the velocity at points B and C because the car is still moving at a constant speed.

At point E, the x position is 20 and the time is 4, so the velocity is (20-15)/(4-3) = 5 units/time. This pattern continues for points F and G, where the velocity remains constant at 5 units/time.

To summarize, the x component of instantaneous velocity at points A-G is 0 at points A and G because the car is not moving horizontally at these points. At points B-F, the x component of instantaneous velocity is 5 units/time because the car is moving horizontally at a constant speed of 5 units per time.