## anonymous one year ago 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?

1. anonymous

@Miracrown

2. anonymous

Excuse me the acceleration is 0.5m/s^2 actually

3. Miracrown

This is a physics Q, right?

4. anonymous

spot on sir

5. anonymous

can also be mathematically computed as well.

6. anonymous

It's hard to tell if a question should belong to physics or maths

7. Miracrown

Aww just how much I love being called ''Sir'' :D

8. Miracrown

$$\color{blue}{\text{Originally Posted by}}$$ @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.

9. Miracrown

I have a question. Does the question mean displacements or positions of 1, 2, and 3 m?

10. Miracrown

Oh, it doesn't give an initial position, so it is displacement. So, can you write or type the quadratic equation you've found?

11. anonymous

Yeah the initial position is 0.00m right on

12. anonymous

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

13. anonymous

d=v(initial)*t+1/2(a)t^2 Where d=displacement or distance v=velocity t=time taken

14. anonymous

It's quid spot on how this equation is supposed to work to compute the distance or displacement.

15. anonymous

However, it's unclear as to whether d stands for displacement or distance in this given situation

16. Miracrown

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.

17. anonymous

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

18. Miracrown

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.

19. anonymous

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

20. anonymous

the vector s is displacement

21. Miracrown

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.

22. anonymous

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

23. anonymous

That's indicated by positive and negative sign here.

24. Miracrown

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

25. anonymous

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.

26. anonymous

That's why I love physics

27. anonymous

relatively less math to deal with but with more intuitive reasoning

28. anonymous

XD

29. Miracrown

Physics is the mother of science