## ParthKohli Group Title OK, here's a nice question that I was just checking out. Try it :-D A particle is at rest and starts to move at a constant acceleration $$\alpha$$. After some time, it starts decelerating at $$\beta$$ and returns to rest. Given that the particle takes time $$T$$ to do so, find in the terms of the known: 1: The highest velocity. 2: The average velocity. one year ago one year ago

1. ParthKohli Group Title

@.Sam.

2. myiah5766 Group Title

The highest velocity.

3. ParthKohli Group Title

Note that the known are $$\alpha, \beta, T$$.

4. .Sam. Group Title

|dw:1368898722315:dw|

5. ParthKohli Group Title

It's a particle, not a cube :-P

6. .Sam. Group Title

Is the whole journey takes time T?

7. ParthKohli Group Title

yes.

8. .Sam. Group Title

|dw:1368898894389:dw|

9. .Sam. Group Title

$\alpha=\frac{v_1-0}{t_1}$ Highest velocity is $$v_1$$ $v_1=\alpha t_1$

10. .Sam. Group Title

$-\beta=\frac{v_2-v_1}{t_2}$ and also $T=t_1+t_2$

11. ParthKohli Group Title

Good going.

12. .Sam. Group Title

Average velocity will be just $\frac{v_1+v_2}{2}?$

13. ParthKohli Group Title

Careful: you don't know the average acceleration as yet.

14. .Sam. Group Title

$a_{ave}=\frac{v_2-v_1}{t_2-t_1}$

15. ParthKohli Group Title

Actually, the average acceleration for the whole thing is $$0$$.

16. .Sam. Group Title

Hmm, lol, but it didn't say it stops at the end and usually when its average acceleration $$\alpha=\beta$$

17. ParthKohli Group Title

Oh, my bad. It does stop at the end.

18. .Sam. Group Title

Hmm $v_2=0$

19. ParthKohli Group Title

What is $$v_1$$? The maximum velocity or the start velocity?

20. .Sam. Group Title

The diagram said it's maximum velocity

21. ParthKohli Group Title

OK. I got confused when you said v1 = v2

22. .Sam. Group Title

Nah

23. .Sam. Group Title

So we'll mix them up $v_1=\alpha t_1 ~~~~~~ -\beta t_2=-v_1$ Getting $\beta t_2= \alpha t_1$

24. ParthKohli Group Title

You just eliminated $$v_1$$ when you actually need it.

25. .Sam. Group Title

Right, $v_1=\frac{\alpha(T-t_2)}{\beta}$

26. ParthKohli Group Title

But $$t_2$$ is not a known thing. You defined it. Express in terms of alpha, beta and T.

27. .Sam. Group Title

28. .Sam. Group Title

/displacement

29. ParthKohli Group Title

You have to calculate that. Both 1 and 2 are to be given in terms of alpha, beta and T.

30. ParthKohli Group Title

Do you want to try more?

31. .Sam. Group Title

I'm getting more variables, lol

32. .Sam. Group Title

But it would've been easier if we used calculus method

33. ParthKohli Group Title

OK, so like $$v_1 = \alpha t_1 \Rightarrow t_1 = \dfrac{v_1}{\alpha}$$. Similarly $$0 = v_1 - \beta t_2 \Rightarrow t_2 = \dfrac{v_1}{\beta}$$. Now$T = t_1 + t_2 = \frac{v_1}{\alpha} + \frac{v_1}{\beta}$Solving for $$v_1$$, we get $$v_1 = \dfrac{\alpha\beta T}{\alpha + \beta }$$

34. .Sam. Group Title

Aw man it was getting close :P

35. ParthKohli Group Title

:-D

36. ParthKohli Group Title

Try the 2nd one maybe?

37. .Sam. Group Title

Average velocity for the whole journey is it?

38. ParthKohli Group Title

Yeah.

39. .Sam. Group Title

is it zero

40. ParthKohli Group Title

No, because the displacement isn't zero.

41. .Sam. Group Title

But the initial and final velocities are zero

42. ParthKohli Group Title

I have not asked for the average acceleration.

43. ParthKohli Group Title

:-P

44. ParthKohli Group Title

$\overline{v} = \dfrac{\Delta s}{\Delta t}$

45. .Sam. Group Title

I've got $\frac{x_2}{x_3-x_2}=\frac{\beta}{\alpha}=\frac{t_1}{t_2}$