## ksaimouli 2 years ago The position of a particle moving along a line is given by

1. ksaimouli

\[s(t)=2t^3-24t^2+90t+7\]

2. ksaimouli

for what values t is the speed of particle increasing

\[s'(t) \implies velocity \space \space s''(t) \implies acceleration\]

4. ksaimouli

ya i got a(t)=0 i got 4

5. ksaimouli

as critical point

6. ceb105

ds/dt=6t^2-48t+90 thus d^2s/dt^2=12t-48 => aceleration is greater than zero when t>4

well first get the turning point, you mean s'(t)=0

8. ksaimouli

no i mean s''(t) gives acceleration =0 will give us information about velocity

yes so you got 4 right

10. ksaimouli

yes

11. ceb105

So yea it is when t>4 that the acceleration is postivei .e. speed increases

yes

13. ksaimouli

no that not the answer i am looking for

14. ceb105

Well its the right one...

15. ksaimouli

but the answer is 3<t<4 and t>5 ( i was shocked !!!!

16. ceb105

is it possible that this is wrong?

17. ksaimouli

@hartnn

|dw:1360515901447:dw| graph increasing the speed inceases when the grapg rises so x<a and x>b

19. ceb105

Okay well 3 and 5 are the critical points of the velocity

|dw:1360516136831:dw|

|dw:1360516160914:dw|

22. ksaimouli

But why i need the critical points of velocity when they ask for velocity

23. ksaimouli

This is no calculator question so i cannot graph

they did not say the rate of change in velocity nor speed just velocity increasing

yeah i know what i am saying is\[s'(t)=0\]

26. ksaimouli

i e 0,3,5

27. ksaimouli

then how did they get 4

so you get 3 and 5 right

29. ksaimouli

yes

speed =|velocity| \[|6t^2-8t+15|>0\] i would like to explain grapghically 4 a sec

i mean\[|t^2-8t+15|\]

32. ksaimouli

its actually velocity=I speedI

33. ksaimouli

velocity is vector where as speed is scalar

i agree with person 3

36. ksaimouli

i found out that they used velocity and acceleration derivatives and they set them =0 and found the answer but i have learned that to find graph of y' use y'' to find max or min

yes so we found \[|t^2-8t+15|>0 \implies t<3 ,t>5\] but also for - velocity we put \[-(t^2-8t+15)\] its positive for \[-(2t-8)>0 \implies t<4\] (3,4) and x>5