1. ## SUVAT(time)

A particle is moving along a straight line OA with constant acceleration 2ms-2. At OA the particle is moving towards A with speed 5.5ms-1. The distance OA is 20m. Find the time the particle takes to move from O to A.

S=20 V=5.5 A=2 T=?

I used the formula $\displaystyle s=vt-1/2at^2$ but I got the wrong answer.

Any help?

2. Originally Posted by Mist
A particle is moving along a straight line OA with constant acceleration 2ms-2. At OA the particle is moving towards A with speed 5.5ms-1. The distance OA is 20m. Find the time the particle takes to move from O to A.

S=20 V=5.5 A=2 T=?

I used the formula $\displaystyle s=vt-1/2at^2$ but I got the wrong answer.

Any help?
I can't see any other method to solve this. In this case I am using A to denote acceleration as to not confuse with a in the quadratic formula

$\displaystyle s = vt + \frac{1}{2}At^2$

$\displaystyle \frac{1}{2}At^2 + vt -s = 0$

$\displaystyle t = \frac{-v\pm \sqrt{v^2+2As}}{A}$

Sub in what you know and I get t = 8

3. Originally Posted by e^(i*pi)
Sub in what you know and I get t = 8

The formula has been rearranged wrong

4. Originally Posted by Mist

The formula has been rearranged wrong
From what I can gather you've took the initial speed, u as the final speed, v. If we define O as the start point and O to A is the positive direction then at point O the speed given is the initial speed.

My equation is fine, I didn't evaluate both roots in my previous post - the second root is 2.5

$\displaystyle s = ut+\frac{1}{2}at^2$ (this is one of the suvat equations)

s = 20, u = 5.5, t = t, a = 2

$\displaystyle 20 = 5.5t+\frac{1}{2}(2)t = 5.5t+t^2$

As this is a quadratic rearrange into the form $\displaystyle ax^2+bx+c=0$

$\displaystyle t^2+5.5t-20 = 0$

Solving that using the quadratic formula gives $\displaystyle t = 8$ and $\displaystyle t= 2.5$.