1. ## journey

A car starts from rest at P and moves with constant acceleration k m/s2.
Three seconds later another car passes through P travelling in the same direction with constant speed u m/s, (i) , show that the second car will just catch up on the first if u = 6k and that it will not catch up on it if u < 6k.

using s=ut+.5at^2 i have formed the 2 equations s=.5kt^2 and s=6kt-18k for the 1st and 2nd car respectively. im not sure what to do next .

2. ## Re: journey

sometimes it helps keep things straight by writing

$s_1(t) = 0.5 k t^2$

$s_2(t) = 6kt - 18k$

i.e. these are both functions of $t$

so at some time $t$ in the future car 2 will catch up (or not) to car 1, i.e. $s_1(t)=s_2(t)$

I'm going to let you finish but what I suspect you'll find is that this equation is true for some real value of $t$ for $u \geq 6k$ and but only for a complex value of $t$ for $u < 6k$

3. ## Re: journey

Originally Posted by markosheehan
A car starts from rest at P and moves with constant acceleration k m/s2.
Three seconds later another car passes through P travelling in the same direction with constant speed u m/s, (i) , show that the second car will just catch up on the first if u = 6k and that it will not catch up on it if u < 6k.

using s=ut+.5at^2 i have formed the 2 equations s=.5kt^2 and s=6kt-18k for the 1st and 2nd car respectively. im not sure what to do next .
Well, another of your algebra drills disguised as a kinematics problem ...

displacement of the accelerating car is $d = \dfrac{k}{2}t^2$ ... in three seconds, it has traveled $d = \dfrac{9k}{2}$ and reaches a speed of $v = 3k$

let $t = 0$ be the time the second car passes point P ...

distance traveled by constant velocity car = head start distance + distance traveled by accelerating car

$ut = \dfrac{9k}{2} + 3kt + \dfrac{k}{2}t^2$

equation is quadratic in $t$, so a re-arrangement ...

$0 = \dfrac{k}{2}t^2 +(3k-u)t + \dfrac{9k}{2}$

multiply every term by 2 to clear the fractions ...

$0 = kt^2 + 2(3k-u)t + 9k$

solve for $t$ using the quadratic formula ...

$t = \dfrac{2(u-3k) \pm \sqrt{4(3k-u)^2 -4k(9k)}}{2k} = \dfrac{2(u-3k) \pm \sqrt{u^2-6ku}}{2k}$

for the time calculation to make sense, the discriminant, $(u^2-6ku)$, has to be $\ge 0$

$u^2 - 6ku \ge 0$

$u(u-6k) \ge 0$

already know that $u > 0$, that means $u - 6k \ge 0 \implies u \ge 6k$

4. ## Re: journey

i am trying to use romseks way. once i get my value for t which is 6 what i do next

5. ## Re: journey

You found a solution for time when u = 6k

You have two more cases to deal with ... u > 6k and u < 6k

Buena suerte.

6. ## Re: journey

So for example will I use 5k and then show that there is no value for t which this works

7. ## Re: journey

You can try that, but how do you know that 5.5k won't work? Are you going to test every single value less than 6k?

8. ## Re: journey

you are right but i cant think of any other way to show it that i understand