Boat deceleration distance using v² = v₁² + 2ad

[Erectable]

Sorry to waste your time on wot should be a simple question:

A boat is traveling at 30 metres per second and begins to decelerate at 0.6 metres per second/per second. How much distance would it cover ( in metres) would it cover before it comes to a compete stop?

I have tried solving this with the question v2^2 = v1^2 + 2ad.

v1 = The intial velocity which is 30 metres per second
v2 = The final velocity which I believe is 0 meters per second; a complete stop.
a = The acceleration which is 0.6 metres per second/per second
d = Distance covered

When I try to solve the equation and isolate the variable "d" I get a rediculous answer.

Am I doing something wrong? Is v2 not 0 metres per second? Is this the wrong equation? Do I need to find out other variables such as time to solve this?

All help is appreciated.

 

[matthew baird]

lets see here...
V^2=Vo^2+2ad
so
0=900m/s+2(-0.6m/s)(d)
-900/2/(-0.6)=750m=d
750meters
I hope this helps, I think what you did was not account for the acceleration being negative since you are slowing down..??

 

[Erectable]

Thanks mate. Initially I tried it with -0.6 m/s/s but still came out with a nonsense answer. My problem it seems is the mathematics of the problem and being able to isolate "d".

I left school a long time ago and have a vague memory of how to solve basic equations, thankyou for clearing the matter up for me. I will now remember that to swap over something that is being multiplied, you divide the entire left hand side by that value.

Well done.