Man has a varying displacements ranging from 1.0m,2.0m and 3.0m from his starting position and he had initial velocity of -2.0m/s and acceleration of 2.0m/s^2. Calculate the time elapsed at times when he is exactly at the 1.0m, 2.0m, 3.0m displacement from his starting point. Formula required. I am actually stuck with the formula that may or may not require quadratic solution to isolate (t). Anyone?

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Man has a varying displacements ranging from 1.0m,2.0m and 3.0m from his starting position and he had initial velocity of -2.0m/s and acceleration of 2.0m/s^2. Calculate the time elapsed at times when he is exactly at the 1.0m, 2.0m, 3.0m displacement from his starting point. Formula required. I am actually stuck with the formula that may or may not require quadratic solution to isolate (t). Anyone?

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

Excuse me the acceleration is 0.5m/s^2 actually
This is a physics Q, right?

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

spot on sir
can also be mathematically computed as well.
It's hard to tell if a question should belong to physics or maths
Aww just how much I love being called ''Sir'' :D
\(\color{blue}{\text{Originally Posted by}}\) @Robert136 It's hard to tell if a question should belong to physics or maths \(\color{blue}{\text{End of Quote}}\) Nah mate, so easy to tell! But in some cases it may not be so easy. Your not wrong.
I have a question. Does the question mean displacements or positions of 1, 2, and 3 m?
Oh, it doesn't give an initial position, so it is displacement. So, can you write or type the quadratic equation you've found?
Yeah the initial position is 0.00m right on
It's supposed to go negative position for a while whilst the initial velocity holds negative but then due to acceleration it's turned to positive and eventually you know what's happening
d=v(initial)*t+1/2(a)t^2 Where d=displacement or distance v=velocity t=time taken
It's quid spot on how this equation is supposed to work to compute the distance or displacement.
However, it's unclear as to whether d stands for displacement or distance in this given situation
This equation is for displacement. So, here's what's awesome about it. You don't have to worry about the fact that the man has an initial negative velocity and will have negative displacements. All you have to do is use d = 1.0, 2.0, and 3.0 m.
So I plug in any of the given values being investigated and leave out the t and t squared, which is bit of a problem because then quadratic formula needs a peek
It's great that you recognize what the guy is actually doing, though! :D Right, you don't substitute anything for t because that's what you're solving for.
Use the equation \[\vec s=\vec ut+\frac{1}{2}\vec a t^2\] As this looks like a question in 1D, you can simply replace vectors with scalars accompanied by appropriate plus or minus sign
the vector s is displacement
d = v0 * t + (1/2)a*t^2 1.0 = -2t + (1/2)(0.5)t^2 0 = -1 - 2t + 0.25t^2 And once you have this, you can solve using the quadratic formula or by graphing.
Here's the deal, as you are dealing in 1 dimension, you can only move forward or backward, like along a line. So you can assign either the forward or the backward direction a positive sign and it's opposite direction as negative so if a quantity is given as positive and another quantity is given as negative, what they are basically saying is they are in the opposite direction
That's indicated by positive and negative sign here.
To be clear, when I say "graph it", I mean you'll input y = -1 - 2t + 0.25t^2 and the find its zeroes and your calculator will use x instead of t
Right. I need to find the roots which would have two obviously but I need to pick the positive one because negative one will be unrealistic to even assume it exists.
That's why I love physics
relatively less math to deal with but with more intuitive reasoning
XD
Physics is the mother of science

Not the answer you are looking for?

Search for more explanations.

Ask your own question