## ParthKohli 2 years ago Classical mechanics.

1. ParthKohli

|dw:1370881868506:dw|At what time will the particles collide?

2. ParthKohli

@amistre64

3. ParthKohli

|dw:1370882089218:dw|

4. ParthKohli

I tried this:$s_1 = 4t + \dfrac{1}{2}t^2$$s_2 = 2t + \dfrac{1}{2}t^2$And $$s_1 + s_2 = 30$$ so$t^2 + 6t = 30 \Rightarrow t^2 + 6t - 30 = 0$

5. ParthKohli

But am I right?

6. amistre64

i see that is what, acceleration up top and speed on the bottom?

7. ParthKohli

Yes.

8. amistre64

$\frac12at^2+v_it+d_i=\frac12\alpha t^2+\beta_it+\delta d_i$ $\frac12t^2+4t+0=-\frac12t^2+2t+30$

9. amistre64

im wondering it that should be a -2t over on the right

10. ParthKohli

That is what I have also been confused about for a while, but I figured that it should be a $$+$$ because we're dealing with distances and not displacements. If we were dealing with displacements, $$s_1 + s_2$$ would not have been $$30$$ either.

11. amistre64

since velocity is a vector, all assume -2t $\frac12t^2+4t=-\frac12t^2-2t+30$ $t^2+6t=30$

12. ParthKohli

Well, so that leads to $$t^2 + 6t - 30$$... so I am right?

13. ParthKohli

Thanks, another one coming up: this time with circles!

14. Vincent-Lyon.Fr

Yes this is correct. Choose solution with t>0.

15. ParthKohli