what would the instantaneous velocity be at 7.50 seconds and how would i solve instantaneous velocity for other values?
edit: solved, thanks for help
var1ance [#3]search it online yw
i searched up how to do it and it's making me draw a tangent line
it's online so i obv can't draw it
Instantaneous velocity is dx/dt which is the slope of the graph.
The answer is the slope of the graph at the required instant (t = 7.5).
slope at 7.5 = (0 - -5)/(8-7) = 5/1 = 5
assuming x is -5 at t=7 and x is 0 at t=8.
It depends on being able to read the value of x precisely from the graph. Assuming x to be -2 at 7.5 is also fine, but you'll get a different answer.
RickyIndian [#11]so -2/7.5 ?
yea but it would +ve since it is moving forward (x coordinate increasing
EntryKJ [#13]Instantaneous velocity is dx/dt which is the slope of the graph.
The answer is the slope of the graph at the required instant (t = 7.5).slope at 7.5 = (0 - -5)/(8-7) = 5/1 = 5
assuming x is -5 at t=7 and x is 0 at t=8.
It depends on being able to read the value of x precisely from the graph. Assuming x to be -2 at 7.5 is also fine, but you'll get a different answer.
i included the graph that the question is referring to and the question is asking what the instantaneous velocity would be at t = 7.50 seconds
RickyIndian [#17]i included the graph that the question is referring to and the question is asking what the instantaneous velocity would be at t = 7.50 seconds
mb, edited
EntryKJ [#13]Instantaneous velocity is dx/dt which is the slope of the graph.
The answer is the slope of the graph at the required instant (t = 7.5).slope at 7.5 = (0 - -5)/(8-7) = 5/1 = 5
assuming x is -5 at t=7 and x is 0 at t=8.
It depends on being able to read the value of x precisely from the graph. Assuming x to be -2 at 7.5 is also fine, but you'll get a different answer.
ofc indian will answer this with detail
BoF7ooM [#16]For instantaneous velocity at 7.5 seconds, find the point at which it is 7.5 second and find its gradient. Here from 7.5 seconds to 8 seconds it is a constant slope, so just find the gradient of that line.
so at 7.5 seconds, the x value is -2 and at 8 seconds the x value is 0.
(0 - - 2) / (8 - 7.5)?
RickyIndian [#23]so at 7.5 seconds, the x value is -2 and at 8 seconds the x value is 0.
(0 - - 2) / (8 - 7.5)?
yeah, it depends on the whether x value is actually -2. It's hard to make out from the graph.
I assumed x to be -5 at t=7, but I can't be sure.
RickyIndian [#23]so at 7.5 seconds, the x value is -2 and at 8 seconds the x value is 0.
(0 - - 2) / (8 - 7.5)?
Yes. Correct.
EntryKJ [#27]yeah, it depends on the whether x value is actually -2. It's hard to make out from the graph.
I assumed x to be -5 at t=7, but I can't be sure.
yup, that worked.
just so i fully understand, if i were to solve for t = 3.0 and t = 4.5, it would go like this
t = 3
(5-10) / (4-2) = -1.25
t = 4.5
(5-5) / (5-4) = 0
RickyIndian [#30]yup, that worked.
just so i fully understand, if i were to solve for t = 3.0 and t = 4.5, it would go like this
t = 3
(5-10) / (4-2) = -1.25t = 4.5
(5-5) / (5-4) = 0
It's correct (edited)
RickyIndian [#32]so what should i change in the calculations?
It depends on which instant you want to find the instantaneous velocity for. If I want to find the instantaneous velocity at 4.5, I would use t=4 and t = 5 since it's a straight line there, so the slope can be found easily. If I want to find the instantaneous velocity at 3, I will use 2 and 4.
RickyIndian [#32]so what should i change in the calculations?
wait, mb I didn't read the calculations fully, you are correct for both. Sorry man, I should have read the equations.
H0r0sha [#35]v=dx/dt since it is a (x vs t) graph , just find the slope of graph at the required time. From img: v= (0-(-2))(8-7.5) = 4
answer is 5 actually
(0--5) / (8-7)
EntryKJ [#34]wait, mb I didn't read the calculations fully, you are correct for both. Sorry man, I should have read the equations.
allg thanks for all your help
RickyIndian [#36]answer is 5 actually
(0--5) / (8-7)
Yeah, then x is not -2 at t=7.5, I think graph should have more lines to figure it out accurately.