# Homework Help: Simple Distance to Time Equation

1. Jul 19, 2011

### physicswonder

I know that s = ut + 0.5at^2 is a simple equation to find distance.
Assuming that initial velocity does not equal zero, how do you find what time equals since my knowledge of quadratics does not seem to be helping.

2. Jul 19, 2011

### cupid.callin

do you have any specific question or i should make one up?

3. Jul 19, 2011

### litup

For some reason, maybe because it was originally in German, we use S=(A*T^2)/2 which is just a little bit different way of writing your equation. So divide by A gives you S/A=T^2/2
Multiply both sides by 2 gives 2S/A=T^2, so (2S/A)^0.5=T.

So in the quarter mile(if you can stomach using feet) it would be 1320 times 2=2640, assume one g, 32 F/S/S, 2640/32=82.5, square root of that is 9.083 seconds to do the quarter mile assuming one g of constant acceleration. So if someone does 9 seconds in the quarter mile, he is averaging 1 g.

And using V=AT, 32 F/S/S * 9.083= 290 Feet per second at the end or times 1.4666 = about 198 miles per hour.

As for the non-zero velocity, make sure it is in the same units and just add the average velocity of acceleration, 1320/9.083=145 feet per second average, plus say, 88 f/s (starting the run at 60 mph) and adding them, average velocity is 233 f/s so 1320/233=~5.6 seconds. So add the average velocities and divide by the distance to give the time with some initial velocity.

Last edited: Jul 19, 2011
4. Jul 20, 2011

### cupid.callin

Lets assume an example: u= -1, s=1.5, a=1

$1.5 = -t + \frac{1}{2}t^2$

$3 = -2t + t^2$

$t^2 - 2t - 3 = 0$

now you have to split the -2t into 2 parts such that their sum is -2t (obvious) and their product is product of the rest 2 terms i.e. (t2)*(-3) = -3t2

i think its -3t,t ... -3t+t = -2t and -3t*t= -3t2

$t^2 - 3t + t - 3 = 0$

$t(t - 3) + (t - 3) = 0$

$(t - 3)*(t + 1) = 0$

$(t - 3) = 0 or (t + 1) = 0$

$t = 3 and t = -1$

but t=-1 makes no sense so t=3 is answer

5. Jul 20, 2011

### litup

Here is a quadratic formula for non-zero starting velocity:

V=initial velocity
A=acceleration
S=distance
T=time to travel the given distance
(-2V+(4V^2+8AS)^1/2)/2A=T

So given units in feet, a car with a running start going 88 feet per second(60 mph) entering the acceleration run of a quarter mile and accelerating at exactly one g (32 f/s/s) takes 6.74 seconds. You can work out the numbers I assume. My earlier answer of averaging velocity was wrong.