Question

vC = 20 - 3aT = √[10²+ 2a(10T + ½aT²)]
what is T equal to?

Answer 1

\(\Large \bf vC=20 - 3aT=\sqrt{10^2+2a\left(10T+\frac{1}{2}aT^2\right)}\quad ?\)

Answer 2

is it \(\large \bf \sqrt{c}=20 - 3aT-\sqrt{10^2+2a\left(10T+\frac{1}{2}aT^2\right)}\quad ?\)

Answer 3

no VC is not involved i want to find t from this 20 - 3aT = √[10²+ 2a(10T + ½aT²)]

Answer 4

\(\large \bf 20 - 3aT=\sqrt{10^2+2a\left(10T+\frac{1}{2}aT^2\right)}\quad ?\)

Answer 5

yes

Answer 6

have you covered the "binomial theorem" yet?

Answer 7

i did but i forgot how to use it

Answer 8

well, you can just raise both sides to the 2nd power, to get rid of the radical
then on the left-side, expand the factors
then on the right-side distribute the "2a"

\(\bf 20 - 3aT=\sqrt{10^2+2a\left(10T+\frac{1}{2}aT^2\right)}\\ \quad \\ \quad \\

(20 - 3aT)^2=\left[\sqrt{10^2+2a\left(10T+\frac{1}{2}aT^2\right)}\right]^2\\ \quad \\
\implies (20 - 3aT)(20 - 3aT)=10^2+2a\left(10T+\frac{1}{2}aT^2\right)\\ \quad \\
\implies (20 - 3aT)(20 - 3aT)=10^2+{\color{blue}{ 2a}}(10T)+{\color{blue}{ 2a}}\left(\frac{1}{2}aT^2\right)\)

Answer 9

expand both sides, then solve for "T"

Answer 10

300-140at+9a^2t^2=at^2