Car accelerating uniformly for 6s - kinematics

[krazykaci]

A car starts from rest and accelerates uniformly for 6s. The distance traveled in the 6s is greater than the distance traveled in the 1st second by a factor of what?

how would i go about doing this?

 

[quasar987]

Use the equation of kinematics at constant acceleration:

$$x(t)=x_0+v_0t+\frac{a}{2}t^2$$

Find x(6) and x(1). Then find the factor k such that x(6)=k*x(1). Makes sense?

 

[EP]

Use your kinematic equation for distance.

 

[krazykaci]

ok I am going to say what i know

v.=0
t=6s
x.=0
a=unknown
but what would final x be?

 

[quasar987]

final x is given by the equation of kinematics.. this is what it says: "give me an initial position, initial velocity and a time and I'll give you a final x"

 

[krazykaci]

but how can i use the equation when i don't know a, or final x and both are in the equation?

 

[EP]

a is constant right? So its just a proportion. The answer will have accelerration in it.