1. ## kinematics problem.

rocket powered sledge accelerates at 35ms-2 and takes 1.6s to pass through a 60m section of a test track. Find the speed at which it entered the section what was its speed when it left?

I think this is the equation to use x=ut + 1/2at2 and than use a=v-u/t to get v ?

a= 35m/s
t= 1.6s
u=?
v=?
x= 60m

thanks

2. Originally Posted by hana_102
rocket powered sledge accelerates at 35ms-2 and takes 1.6s to pass through a 60m section of a test track. Find the speed at which it entered the section what was its speed when it left?

I think this is the equation to use x=ut + 1/2at2 and than use a=v-u/t to get v ?

a= 35m/s
t= 1.6s
u=?
v=?
x= 60m

thanks
You are correct, but let me clean this up a bit.

Let the start of the track be your origin and let the +x direction be the direction of the acceleration.

So as you suggested
$x = x_0 + v_0t + \frac{1}{2}at^2$
becomes
$x = v_0t + \frac{1}{2}at^2$
and gives you a $v_0$.

Then $v = v_0 + at$

-Dan

3. When you manipulate the formula to give ‘v’ it does not equal 9.5.
u= 1/2at2 + x / t
??

4. Originally Posted by topsquark
You are correct, but let me clean this up a bit.

Let the start of the track be your origin and let the +x direction be the direction of the acceleration.

So as you suggested
$x = x_0 + v_0t + \frac{1}{2}at^2$
becomes
$x = v_0t + \frac{1}{2}at^2$
and gives you a $v_0$.

Then $v = v_0 + at$

-Dan
You forgot to divide the acceleration term by t when you solved for v, and you have some negation problems:
$u = \frac{x}{t} - \frac{1}{2}at$

Okay, let's do some numbers:
$x = 60~m$
$a = 35~m/s^2$
$t = 1.6~s$

So
$60 = 1.6v_0 + \frac{1}{2}(35)(1.6)^2$

Gives
$v_0 = 9.5~m/s$

and
$v = 9.5 + (35)(1.6) = 65.5~m/s$

-Dan