Calculating Velocity and Time Using Newton's Second Law

[404]

A sled of 6.0 kg mass is moving along a smooth, horizontal ice surface with a velocity of X. A force of 36N is applied to the sled in its direction of motion, increasing its velocity to 2X while it moves 10m. Find the sled's original velocity, x. And the length of time that the force acted.

Smooth surface means frictionless, and the answers should be 6.3m/s and 1.1 sec respectively. Thanks.

 

[what]

Here are the eqs that you would use
f =ma
Vf= Vi +at
d= Vi(t) + .5a(t^2)

 

[404]

is there a way to solve it w/o d= Vi(t) + .5a(t^2)?

Our teacher forbid us to use it. We can only use the basic ones like a=delta v/ delta t, v average = delta d / delta t.

 

[whozum]

Why don't you tell us what you think so we can tell you what we think of what you think.

 

[what]

what whozum said, you have to show us your work for us to figure out what you did wrong or right. It would be unfair if we just gave you the answer, but those eqs work as well.

 

[404]

Well there's like 20 questions in total, this is the only one out of the bunch I didn't get. I tried to average velocity formula

like...

I found acceleration is 6 m/s^2, by using 36N/6kg

v average = (x + 2x)/2 So velocity average is 1.5x, and I put it back into the velocity average = d / t formula, since i know d, I found t to be 6.67 s, then plugged it into the acceleration formula, 6 = delta v / 6.67 (time) and basically the embarrassing result is nothing close to the answer...

That was only the first part, never tried to find time as this answer is not correct.

 

[whozum]

Do you knwo the equation for final velocity without knowing time? It involves just acceleration distance and velocity.

 

[what]

Your acceleration is correct.
t=6.67 s is not correct
So you put 1.5x = 10/t and got t= 6.67s where from?
You have two variables therefore t= 10/(1.5x)
now plug that t into your acceleration formula and you will get your speed.

 

[404]

Well we are forbiden to use those equations, only the simple ones.

And I did plug it in, the variables simpled canceled out for me. I tried 6 = (2x-x)/t
t=(2x-x)/6 so i set that equal to (2x-x)/6 = 10/1.5x... (I'm still working at the question though...)

 

[what]

(2x-x)/6 = 10/1.5x is the right eq. just a matter of algebra now.

 

[404]

If I'm not mistaken, that comes to

x/6 = 10/1.5x

The x's would cancel...

 

[whozum]

Nope, multiply both sides by 1.5x

 

[404]

Lol...Oh jeeze. My bad, I don't know what gotten over me, SO obvious, yet I somehow missed it.

EDIT: Got both answers now, thanks guys.