Chuck's Methods Questions

[chuckjefster90]

Having sumore trouble 

essential maths 12 j q. 7

a body starts from O and moves in a straight line. After t seconds (t>0) its velocity v cm /s is given by v= 2t-3

d) find the distance travelled in first 3 seconds

e) find its average speed for first 3 seconds

any help wud be appreciated

[khalil]

Ok this is trickly cos' the graph has area above and below the graph
If you graph v= 2t-3 you'll see that between x=0 and x=3 there are two parts to the area under and above the graph
Our job is to find the area under and above the graph, this will tell us the DISTANCE travelled
There is a pattern derive original to get velocity, derive velocity to get acceleration
Anyway, all we do is get the area under the graph and it is done in two parts.
|| + where v=2t-3 and a=1.5

I got 1.5 by make v=0, so I could get its intercept
The first integratinon sum is +ve cos' area cannot be negative

e) distance/time= 4.5/3=1.5

[chuckjefster90]

thnx

also q9

A body moves in a straight line with an acceleration of 8m/s. If after 1 second it passes through O and after 3 seconds is 30 metres from O.

What is its intitial displacement relative to O?

[khalil]

As I said before: antiderive acceleration to get velocity and antiderive velocity to get distance.
We are given acceleration: 8
= 8t+c
= 4t^2+c+k

We know have the distance formula after anti deriving twice d= 4t^2+ct+k
Sub in the known values (1,0) and (3,30) into d and solve for t and k.
Once done we have our new distance formula d=4t^2-t-3
To find initial displacement sub in t=0. Thus initial displacement is -3m

[chuckjefster90]

Can someone please explain the inverse properties of logs and how they work?
loga(a^x) = x
a^logax=x

i dont really get it

[chuckjefster90]

also

When depth of liquid in container is xcm the volume is x(x^2 +36) cm 3. Liquid is added to the container at a rate of 3 cm3/ s
Find the rate of change of depth of liquid at the instant when x=11

[Flaming_Arrow]

dV/dt = 3

dx/dt = dV/dt * dx/dV

dV/dx = 3x^2 + 36

dV/dx (11) = 399

dx/dt = 3 * 1/ 399 = 1/ 133 cm/s

[TrueTears]

Can someone please explain the inverse properties of logs and how they work?
loga(a^x) = x
a^logax=x

i dont really get it

Let

Thus,

But

so

[chuckjefster90]

a^{log_ax}

Let log_ax = y

Thus, a^y = x

But y = log_ax

so a^{log_ax} = x


i still dont get this inverse property, how does it equal x?

[TrueTears]

That was just a proof.



Now let

Now just focus on [Look I could have done let would be the same]

So , now [Just simple log here]

Now look at , what did we say y was? We let , sub this back in and you get

QED.

[chuckjefster90]

sweet thanks