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1. I need help solving tis question, I know which formula to use but I just don't get the correct answer on the calculator.

Here's the question:

A particle is moving along a straight line OA with constant acceleration 2.5ms-1. At time t=0, the particle passes through 0 with speed 8ms-1 and is moving in the direction OA. The distance OA is 40m. Find the time taken for the particle to move from O to A.

The formula I used is S=vt-1/2at^2 and I got the wrong answer.

Any help?
2. been a while since i did any of this but might help if you post what answer you get so i can try to see where your going wrong
3. Its fairly simple
Use SUVAT since its constant acceleration
s=40m
U=8m/s
V
a=2.5m/s^2
t=?
Sub numbers into
s=ut + 1/2at^2
rearrange to get a quadratic and solve using quadratic equation or complete the square
4. The formula is s=ut+1/2at^2

But, rather than using that formula and generating a very annoying quadriatic, it might be easier to use:

Use v = sqrt (u^2 + 2as)
v = sqrt (64 + 2*2.5*40)
v = 16.248

Then use t = (v-u)/a
t = 16.248-8 / 2.5
t = 3.3s

If you wanted to use the first formula, do

s = ut + 1/2 a t^2
40 = 8t + 1.25t^2
0 = 1.25t^2 + 8t - 40
0 = 5t^2 + 32t - 40

Use Quadriatic formula now (see why I suggested against this method?) to generate:

t = -9.7s or 3.3s
t = -9.7s or 3.3s (NEGATIVE VALUE IS INVALID)
t = 3.3s
5. Have you tried: s = ut + {1/2}at² ?
6. (Original post by innerhollow)
The formula is s=ut+1/2at^2

But, rather than using that formula and generating a pain of a quadriatic, it might be easier to use:

Use v = sqrt (u^2 + 2as)
v = sqrt (64 + 2*2.5*40)
v = 16.248

Then use t = (v-u)/a
t = 16.248-8 / 2.5
t = 3.3s
Thanks, I see where I went wrong .. i'll rep you tomorrow

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