your answer is might be wrong 460
I think.
2 over
3a3 + 3a2 - 3
You say that there is uniform acceleration so:
vf-vi=at (final velocity minus initial velocity is equal to acceleration times time)
We know vf, vi, and t so we can solve for acceleration:
24-12=a10
12=10a
a=1.2
That is the acceleration, we will need to integrate with respect to time twice...
v=⌠a dt
v=at+vi , we know a=1.2m/s^2 and vi=12m/s
v=1.2t+12,
x=⌠1.2t+12 dt
x=1.2t^2/2+12t+xo, we can just let xo=0 for this problem...
x(t)=0.6t^2+12t
Now we know that this acceleration lasts for 10 seconds so the distance traveled in that time is:
x(10)=0.6(10^2)+12(10)
x(10)=60+120
x(10)=180 meters
I believe that it is a concurrent power
<h3>Given</h3>
P = 3r + 2s
<h3>Find</h3>
the corresponding equation for s
<h3>Solution</h3>
First of all, look at how this is evaluated in terms of what happens to a value for s.
- s is multiplied by 2
- 3r is added to that product
To solve for s, you undo these operations in reverse order. The "undo" for addition is adding the opposite. The "undo" for multiplication is division (or multiplication by the reciprocal).
... P = 3r + 2s . . . . . . starting equation
... P - 3r = 2s . . . . . . -3r is added to both sides to undo addition of 3r
... (P -3r)/2 = s . . . . . both sides are divided by 2 to undo the multiplication
Note that the division is of everything on both sides of the equation. That is why we need to add parentheses around the expression that was on the left—so the whole thing gets divided by 2.
Your solution is ...
... s = (P - 3r)/2
