1. ## journey

a particle starts from rest from A and moves to B with uniform acceleration . in the last 2 seconds of its motion it describes seven sixteenths of the whole distance and in the first second is describes 3m.find the distance from A to B and how long it took to travel to A to B.
for the first second 3=.5a so a =6 and i also formed the equation 12+2u=6/17 s where s stands for total distance . i formed this equation through the area of a trapezium . dont know where to go next

2. ## Re: journey Originally Posted by markosheehan a particle starts from rest from A and moves to B with uniform acceleration . in the last 2 seconds of its motion is describes seven sixteenths of the whole distance and in the first second is describes 3.find the distance from a to b and how long it took to travel to a to b.
for the first second 3=.5a so a =6 and i also formed the equation 12+2u=6/17 s where s stands for total distance . i formed this equation through the area of a trapezium . dont know where to go next
What does this clause mean?
in the first second is describes 3
Does the number 3 have units?

I'm sure you know how to write legible, correctly formatted sentences ... like starting the first word with a capital letter and ending with a period, and inserting a space before the next sentence. Try doing that to effectively communicate what you're describing/asking.

Also, if A & B are points of reference, continue to describe those points with capital letters ... do not change to lower case.

3. ## Re: journey

sorry i have edited it to the exact question

4. ## Re: journey Originally Posted by markosheehan A particle starts from rest from A and moves to B with uniform acceleration . In the last 2 seconds of its motion it describes seven sixteenths of the whole distance and in the first second it travels 3m. Find the distance from A to B and how long it took to travel to A to B.
Let $t$ be the time it takes to travel the entire distance $d$ ...

in the first second it travels 3m
$\dfrac{1}{2}a(1^2) = 3 \implies a = 6 \, m/s^2$

Total distance traveled ...

$d = \dfrac{1}{2}at^2 = 3t^2$

If the particle moves 7/16 of the total distance in the last two seconds, then it travels 9/16 of the total distance in the first $(t-2)$ seconds ..

$\dfrac{9}{16} \cdot d = 3(t-2)^2$

You now have two equations with two unknowns ($t$ and $d$) ... solve the system algebraically.