1. ## uniform acceleration question

A particle X travels with a constant velocity of $\displaystyle 6ms^{-1}$ along a straight line. It passes a particle Y which is stationarity. One second later particle Y accelerates at $\displaystyle 2ms^{-2}$ . How long after being passed it take for particle Y to draw level with the particle X? Leave your answer in surd form.

for particle 'x' $\displaystyle u = 6ms^{-1}$ ,$\displaystyle a = 0ms^{-2}$
t= T seconds.

and for particle 'y' $\displaystyle u= 0ms^{-1}$ , $\displaystyle a = 2ms^{-2}$ and t= T-1 seconds

I am pretty sure you use this formula $\displaystyle s=ut + \frac{1}{2} at^{2}$

but I am not sure how to put all this info together and make one equation any help ?

thanks!

2. Originally Posted by Tweety
A particle X travels with a constant velocity of $\displaystyle 6ms^{-1}$ along a straight line. It passes a particle Y which is stationarity. One second later particle Y accelerates at $\displaystyle 2ms^{-2}$ . How long after being passed it take for particle Y to draw level with the particle X? Leave your answer in surd form.

for particle 'x' $\displaystyle u = 6ms^{-1}$ ,$\displaystyle a = 0ms^{-2}$
t= T seconds.

and for particle 'y' $\displaystyle u= 0ms^{-1}$ , $\displaystyle a = 2ms^{-2}$ and t= T-1 seconds

I am pretty sure you use this formula $\displaystyle s=ut + \frac{1}{2} at^{2}$

but I am not sure how to put all this info together and make one equation any help ?

thanks!
Take $\displaystyle t=0$ to refer to the time when the particles are level.

Then the position of X at time $\displaystyle t>0$ is:

$\displaystyle x(t)=6t$

and of particle Y at time $\displaystyle t>1$:

$\displaystyle y(t)=\frac{2 (t-1)^2}{2}=(t-1)^2$.

When they draw level:

$\displaystyle x(t)=y(t)\ \ \ \ \ \ ..(1)$

and $\displaystyle t>1$.

Now $\displaystyle (1)$ is a quadratic in $\displaystyle t$ and you want the root which is $\displaystyle >1$

.

3. Originally Posted by Constatine11
Take $\displaystyle t=0$ to refer to the time when the particles are level.

Then the position of X at time $\displaystyle t>0$ is:

$\displaystyle x(t)=6t$

and of particle Y at time $\displaystyle t>1$:

$\displaystyle y(t)=\frac{2 (t-1)^2}{2}=(t-1)^2$.

When they draw level:

$\displaystyle x(t)=y(t)\ \ \ \ \ \ ..(1)$

and $\displaystyle t>1$.

Now $\displaystyle (1)$ is a quadratic in $\displaystyle t$ and you want the root which is $\displaystyle >1$

.
I still don't understand how to solve for 't'.

4. Originally Posted by Tweety
I still don't understand how to solve for 't'.
Have you put x(t)=y(t) and expanded? The you have a quadratic in t. Do you know the quadratic formula or the method of completing the square? Then use it.

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