Question

Can anyone help me find time when acceleration, distance and initial velocity is given? Pls!

Answer 1

v = m/s
a = m/s^s
d = m

Answer 2

m = meters
s = seconds

Answer 3

It's one of the kinematic equations in physics\[s=ut+\frac{ 1 }{ 2 }at ^{2}\]
distance=initial velocity times time plus half acceleration times time squared. That's a quadratic in t for which there's another formula ...
http://perendis.webs.com

Answer 4

You can also derive all the kinematic equations from a = dv/dt and v = dx/dt and a = dv/dx

Answer 5

um.. but I don't know what is t... I wanna know the equation for finding time.

Answer 6

For a general quadratic \[ax ^{2}+bx+c=0\]
\[x=[-b \pm \sqrt{b ^{2}-4ac}]/2a\]

Answer 7

for x in my last post, read t in the equation. The coefficients/multipliers are as in the other equation I've posted. s is distance, u is initial velocity, a is acceleration, t is time.

Answer 8

Note also that in that awkward looking "square root" equation, the entire right hand side is over 2a.

Answer 9

There are several kinematic equations. It depends what the problem states since several of the equations have time.

Answer 10

@raffle_snaffle and @logo
True, and logo's problem said time, acceleration, distance and initial velocity. I think that makes it the one that results in the quadratic in time.
In terms of arithmetic, there's a square root in the possible solutions to the quadratic. That says plus or minus values are both "arithmetically/algebraically" possible. So, to get round that, it's the plus value, assuming that clocks and things don't (yet) run backwards.
The good old kinematic equations strike again. Where would we be without them. Smiling more, maybe ?

Answer 11

ps bit more on
http://perendis.webs.com