# 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

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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?

PhanthomJay
Homework Helper
Gold Member
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?

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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!