What Is the Maximum Speed to Stop Safely Within 40 Meters?

[NMiller5a]

I've been trying for about an hour to figure this one out and decided that I should probably ask some people who know what their doing for some input.

"A driver of a car suddenly sees the lights of a barrier 40 m ahead; it takes the drive 0.75 s to apply the brakes; the car accelerates at a speed of -10 m/s^2. What is the maximum speed possible?"

I decided to break the problems into two parts; first, the total distance covered during the time prior to initiating of braking and the time after. For the former part I was left with (Vo)(0.75). The latter part I ended with Vo(Vo/10)+0.5(-10)(Vo)^2. I put those together and 40=(Vo)(0.75)+Vo(Vo/10)+0.5(-10)(Vo)^2. Did I create an incorrect equation?

 

[FZ+]

Welcome to PF!

I think the second part of the equation is incorrect.

You are trying to use s = ut + 0.5 a t^2, right?

You shouldn't as that is 1 equation with two unknowns - you don't know what s is, and you don't know what t is.

Instead, use v^2 = u^2 + 2*a*s, and seen where you get from there.

PS: You know SUVAT terminology, right?
s = displacement
u = initial velocity
v = final velocity
a = acceleration
t = time

 

[NMiller5a]

Thank you for your prompt reply and assistance!

I changed my equation into v=u^2+2(a)(s) and it worked very well. 0=u^2+2(-10)(40-0.75u)] worked perfectly and I got an answer of 21.76 m/s, or about 78 km/h.

When the time comes that I'm only 40 meters away from an inescapable barrier I'll know if I'm doomed or not...