Simple maths question?

Question

For the formula S= ut + 1/2 at squared, find the values of t if:
S=18, u=7, a=2

Please tell me step by step on how to find t please?? 10 points?

That makes total sense thx heaps. I tried to do the next one with same formula.
S=6, u=11, a=4

6=11t + (1/2) 4t (squared)
0=11t + 2t (squared) - 6
0=2t (squared) + 11t - 6
0=t (squared) + 11t - 12
0= (t-1)(t+12)
t=1, -12 therefore 1...

But answer is 1/2, -6?? Help again please?? Im trying..

Answer 1

Substitute the values in the equation:

18 = 7t + .5*2*(t^2)
t^2 + 7t = 18

t^2 + 7t - 18 = 0

The equation has assumed the form of a quadratic equation.

So, we can solve it by mid-term breaking:

t^2 + 9t - 2t - 18 = 0
t(t + 9) -2(t + 9) = 0
(t - 2)(t + 9) = 0

t - 2 = 0 OR t + 9 = 0
t = 2 OR t = -9

Since, time CANNOT be negative, so t = 2

Answer 2

S= ut + ½at²
18 = 7t + ½*2*t²
18 = 7t + t²
t² + 7t -18 = 0
t² +9t - 2t - 18 = 0
t(t+9) -2(t+9) = 0
(t - 2) (t + 9) = 0
Let (t-2)=0, then, t = +2
Let (t+9)=0, then, t = -9
As time can not be -ve, the only value of "t" is +2.
======================================

Answer 3

We have s=18, u=7, a=2
now substituting corresponding valves in the formula
s=ut+1/2at square
==> 18=7 t +1/2 2 t square
= 7t+tsquare
0=tsquare+7t-18
0=tsquare+9t-2t-18
0=t(t+9)-2(t+9)
0=(t+9)(t-2)
==> (t+9)=0 or (t-2)=0

Therefore t=-9 or t=+2

Answer 4

As we know,

s = ut + (1/2) at^2

Directly put the values of s, u and a in the above equation,

18 = 7t + (1/2)*2 t^2
or, 18 = 7t + t^2
or, t^2 + 7t -18 = 0
or, t^2 + 9t - 2t -18 = 0
or, t (t + 9) - 2 (t + 9) = 0
or, (t + 9) (t - 2) = 0

Either (t + 9) = 0 or, (t - 2) = 0

So, t = -9 or t = 2

But time cannot be negative.
So, the one and only value is t = 2

[Answer] t=2

Answer 5

S= ut + 1/2 at^2
putting s=18 ,u=7, a=2
18=7t+1/2*2*t^2
18=t^2+7t
t^2+7t-18=0............(1)
this quadratic
you can use quadratic formula or also by factorization


lets do first by factorization
t^2+7t-18=0
t^2+9t-2t-18=0
taking commons
t(t+9)-2(t+9)=0
(t+9)(t-2)=0
t=-9 if t is time then t=-9 is not possible otherwise tmay be -9
t=2






by quadratic formula
t^2+7t-18=0
t=(-b+-(b^2-4ac)^(1/2))/2a
t=(-7+-(49-4(1)(-18)))^(1/2)/2
t=(-7+-11)/2
t=4/2,-18/2
t=2,-9





hope you will get it

Answer 6

S = (ut+at^2/2)
now just put the given values in the equation:
18=7t+(2/2)t^2 then 18=7t+t^2
moving the constant 18, to the left side we have:
t^2 +7t-18 = 0
to solve, we employ the following equation:
t(or any other variable)=[-b±sqrt(b^2-4ac)]/2a
for the general case at^2+bt+c=0
so we identify and solve:
t=[-7±sqrt(7^2-4x1x(-18))]/2x1 = -7±sqrt(121)/2 = -7±11/2 =
-18/2 or 4/2
we take the second because time is not defined for negative values,so t=2.
If t is not time, the you obtain two values: -9 or 2 (it's true, I'm too used to identify t as time, thx niki einstien).

Answer 7

By substituting the given values we get
t(sqaure)+7t-18=0 which is a quadratic equation, which can be solved in two ways firstly, either by Substitution method or by applying the formula {-b+/-[(b(square)-4*a*c)sq root]} /2a ,
where a=1, b=7 and c= -18 from the above quadratic eqn.
You will get two values of t, one will be positive the other will be negative, consider whichever is relavent as per the given problem.

Answer 8

18=7*t + 0.5 * 2 * t^2
18=t^2+7*t
t^2+7*t-18=0
now apply the formula: t1,t2=(-7+-square root(7^2-4*1*(-18)))/(2*1)
you get t1,t2=(-7+-square root(49+72))/2=(-7+-11)/2
so you have two solutions: t1=(-7+11)/2=2, t2=(-7-11)/2=-9

Answer 9

come on Dylan,
you only have to find one value, the value of t
it is an assignment
begin by substituting the given values, then performe the arithmetic.