# Super congfused about finding distance with constant acceleration

*confused

A car speeding up from rest to constant accel. travels 100 meters in 6.04s. How far did it travel in the first 2.37 seconds?

So I figure if I divide distance by time 100/6.04 it will give me the velocity 16.5m/s. And since distance = vt I would do d=15.6*2.37 which equals 37.03m but this is wrong. I was also told to try drawing a graph of velocity vs time....but how can I graph numbers such as these (high and low #'s with decimals) If I cant graph it I cant find the slope or y intercept....so I'm lost

Last edited:

Related Introductory Physics Homework Help News on Phys.org
The car is accelerating at a constant rate (if I understand the problem correctly, it's wording is strange), so its velocity is constantly changing.

Do you know any kinematic equations for motion under constant acceleration?

In class we've used these: v=v0+at, x=x0+v0t+1/2at^2, and v^2=V0+2a(x-x0)
but from what I see I have an intitial velocity of 0, a distance and starting point which would be x0=0 and x=100m then I have two times..... I'm having trouble fitting this all in..

CAF123
Gold Member
Start by using $x - x_o = v_{o}t + \frac{1}{2}at^2$ to find this acceleration, which is a constant value.

Can I use the eqtn v=v0+at instead with the initial velocity being zero, velocity being 100m/6.04s and t=6.04s? and solving for a?

and them plug it into the x-x0 eqtn?

Last edited:
lewando
Homework Helper
Gold Member
Can I use the eqtn v=v0+at instead with the initial velocity being zero, velocity being 100m/6.04s and t=6.04s? and solving for a?
No. The v in that equation is final velocity. You would be using average velocity. Do what CAF123 suggests.

Ok so when I solve for a the eqtn is a=2x/t^2. Then I plug a back into the eqtn but I use the second given time.... Am I on the right track?

CAF123
Gold Member
Ok so when I solve for a the eqtn is a=2x/t^2. Then I plug a back into the eqtn but I use the second given time.... Am I on the right track?
Yes!