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 $s=vt-1/2at^2$ 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 $s=vt-1/2at^2$ 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

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

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

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

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
Thats the wrong answer, the answer is suppossed to be 2.5

The formula has been rearranged wrong

4. Originally Posted by Mist
Thats the wrong answer, the answer is suppossed to be 2.5

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

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

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

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

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

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

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