Question

Determine the velocity and acceleration of an object that moves along a straight line in such a way that its position is s(t)=t^2+(2t-3)^1/2

Answer 1

Note that \[v = \frac{ ds(t) }{ dt }~~~\text{and}~~~a = \frac{ d^2s(t) }{ dt^2 } = \frac{ dv(t) }{ dt }\]
so given these definitions, what would the velocity and acceleration be?

Answer 2

v(t)=s(t)=2t+1/2(2t-3)^-1/2
a(t)=s1(t)=s2(t)= 2-(2t-3)-3/2

Answer 3

Close, but remember you must apply the chain rule as well, \[\frac{ ds }{ dt } = 2t+\frac{ 1 }{ 2 } (2t-3)^{-1/2}*2\]

Answer 4

wait where you getting that two from

Answer 5

oh ok

Answer 6

Everything clear/

Answer 7

goted

Answer 8

Alright :)

Answer 9

yeah

Answer 10

can you plz help me with one more question

Answer 11

Sure, I'll try

Answer 12

ok wait a sec

Answer 13

Determine the maximum and minimum of each function on the given intervals.
A) F(x)= 2x^3-9x^2, -2<x<4
B) F(x)= 12x-x^3, xzE [-3,5]
C) F(x)= 2x+18/x, 1<x<5

Answer 14

i wanna know how can we find the max and min in this type of situation

Answer 15

First do the derivative, to find the critical points (a,b), so where would the derivative equal to 0 and is undefined, that will give you the critical points.
Then you take the values and see if they are in the given interval, if they are you can continue and plug the values back into your original equation f(x) which then will give you the maximum and minimum values. Lowest value will be the minimum, highest value the maximum.
I hope that was easy to follow.

Answer 16

I find the 0=6x(x-3) so the critical point be 0 and 3.

Answer 17

@iambatman

Answer 18

@nincompoop

Answer 19

yes