Solving Force and Kinematics: Braking Distance of a 760kg Car

[rudebwoi]

hi, could some1 tell me how to solve this question..

a 760 kg car traveling at 90 km/h can brake with a force of 2200 N. How far will it travel in a potential accident situation if the driver's reaction time is 0.55s?

i think u use fnet = ma and one of the kinematics eqns...help please

 

[neutrino]

Welcome rudebwoi.

As a general rule, if you don't know which equation to use "at a glance," then pick the one for which most information has been provided

Let's look at the all the "givens" - acceleration (you know the force on the car and it's mass), initial and final velocities are provided, and the time interval is stated as 0.55s. Which equation connects all this and the distance travelled?

 

[radou]

Which is the equation of displacement for motion with constant acceleration?

Edit: oops.

 

[rudebwoi]

the answer is 135 m...im not getting it

 

[P3X-018]

How do you describe the position (in 1-dimension) of the car as a function of time when you know (or can calculate) the acceleration of the car and you know the initial velocity of the car.

 

[rudebwoi]

:S...not sure

 

[P3X-018]

If the acceleration on the car is $a$ and the car has the initial velocity $v_0$, setting the "start" position to $x_0=0$, then the position as a function of time is

$$x(t) = \frac{at^2}{2}+v_0t$$

EDIT:
Combining this equation with

$$\dot{x}(t) = at+v_0$$

Using that the velocity at the "end" is $v_f=0$, should give you that

$$2a = \frac{-v_i^2}{\Delta x}$$

where $\Delta x$ is the distance travelled, and $v_i$ is the initial velocity.