Distance Traveled in 8s: Factor of 1st Second

[kimikims]

A car starts from rest and accelerates uniformly for 8 s.

The distance traveled in the 8 s is greater than the distance traveled in the first second by a factor of:


----

I think you use this formula, but I don't know.

Vo = 0
X1 = 1/2at1^2
X2 = 1/2at2^2

X1/X2 = (1/2)at1^2 / (1/2)at2^2

= t1^2 / t2^2

 

[marlon]

A car starts from rest and accelerates uniformly for 8 s.

The distance traveled in the 8 s is greater than the distance traveled in the first second by a factor of:


----

I think you use this formula, but I don't know.

Vo = 0
X1 = 1/2at1^2
X2 = 1/2at2^2

X1/X2 = (1/2)at1^2 / (1/2)at2^2

= t1^2 / t2^2

seems ok to me
t1 = 8 and t2 = 1
marlon

 

[M98Ranger]

Since the car is accelerating uniformly, the time intervals (t1 and t2) are the same, so the ratio becomes:

= 1^2 / 1^2

= 1

Therefore, the distance traveled in the first second is equal to the distance traveled in the 8 seconds. There is no factor or difference between them. This is because the car's acceleration is constant, meaning it covers equal distances in equal time intervals.