Help with instantaneous velocity

AI Thread Summary
To find the instantaneous velocity at T=1.4 from the provided graph, one must calculate the slope of the line representing distance over time. The formula for instantaneous velocity is indeed dx/dt, where dx is the change in distance and dt is the change in time. Since the graph is a straight line, the acceleration is constant, making the instantaneous velocity equal to the average velocity over the interval. To compute this, create a table with initial and final points from the graph, and apply the formula at t2 = 1.4 seconds. The discussion clarifies the distinction between instantaneous velocity and acceleration, emphasizing the correct approach to solve the problem.
ramin86
Messages
42
Reaction score
0
I was given the following graph:

http://www.webassign.net/pse/p2-03.gif

I must find the instantaneous velocity at T=1.4

The formula for instantaneous velocity is delta x/delta t as t approaches zero. However, I'm still not sure how to work this problem. Please help
 
Physics news on Phys.org
Find the equation for x(t) in the interval t = [0,2]. You should know the velocity by then.
 
The formula you gave is for instantaneous acceleration, not velocity. So I will assume you are looking for instantaneous acceleration.
You will note the graph is a straight line from 0 - 2 seconds which means the acceleration is constant. The instantaneous acceleration is therefore the same as the average acceleration over that interval.
 
No, its instantaneous velocity that I'm looking for. I thought instantaneous acceleration is the change in velocity over the change in time.
 
Chronos said:
The formula you gave is for instantaneous acceleration, not velocity. So I will assume you are looking for instantaneous acceleration.
You will note the graph is a straight line from 0 - 2 seconds which means the acceleration is constant. The instantaneous acceleration is therefore the same as the average acceleration over that interval.

HOWEVER though, the graph does not show change in velocity over change in time. It's simply distance vs time. Besides, Ramin gave the right formula, distance over change in time. i think you should ignore chrono's comments Ramin...

Since it is a straight line though, the velocity is not changing, so you can just find the instantaneous velocity by finding the slope of the 0-2 second interval(since t = 1.4 has the same velocity as that interval)
 
Last edited:
My bad, you are correct, dx/dt. Make a table x1, t1, x2, t2. x1 and t1 will always be 0,0. x2 and t2 will be per the graph. solve for dx/dt at t2 = 1.4 seconds. Hint dx is always x2 - x1 and dt is always t2 - t1
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top