Calculate the total stopping distance of the car

[eefje]

Homework Statement

reaction time of a driver to brake is 0,5s in that time the speed is constant
the speed of a car is constant when he decides to slow down and to stop with -6,0 m/s2, the initial speed is 50 km/h
Calculate the total stopping distance of the car

Homework Equations

Ve= -6,0*t +13,89
t=2,3
xe=-6,0*(2,3)^2 +13,89*2,3
what did I do wrong?

The Attempt at a Solution

the solution has to be 23,03 m but I can't solve it,

 

[gneill]

Hi eefje, Welcome to Physics Forums.

FYI, you should list the standard equations that you will use in the relevant equations section, then show how you applied them in your attempt at a solution.

I see that you didn't account for the distance traveled during the driver's reaction time period. How far does he travel before he bgins to decelerate?

 

[eefje]

Hi eefje, Welcome to Physics Forums.

FYI, you should list the standard equations that you will use in the relevant equations section, then show how you applied them in your attempt at a solution.

I see that you didn't account for the distance traveled during the driver's reaction time period. How far does he travel before he bgins to decelerate?

Hey Thank you for helping me, I will do it right the next time.
In the 0,5 s he travels 6,945m i think (13,89*0,5). But I still can't solve it

 

[gneill]

What basic equation are you applying for the period in which he decelerates?

 

[eefje]

What basic equation are you applying for the period in which he decelerates?

you mean in the 0,5s ? Δv= Δx/ Δt

and after that: Δx= a*Δt^2+Δv*t
Δv=a*Δt

 

[gneill]

you mean in the 0,5s ? Δv= Δx/ Δt

and after that: Δx= a*Δt^2+Δv*t
Δv=a*Δt

You're good for the first period (reaction time period). $d = vt$

For the second period where acceleration is happening you want the standard kinematic formula $d = v_o t + \frac{1}{2} a t^2$.
Note the "1/2" constant that multiplies the acceleration term.

 

[eefje]

You're good for the first period (reaction time period). $d = vt$

For the second period where acceleration is happening you want the standard kinematic formula $d = v_o t + \frac{1}{2} a t^2$.
Note the "1/2" constant that multiplies the acceleration term.

Thank you very much, I can solve it now. the 1/2 was a stupid mistake, but I didn't know that the distance in the 0,5 s important was. Thank you :)

 

[gneill]

You're welcome. Good luck in your studies.