[sleepyheadzzz]

I can do some of this kinematics stuff, but I don't understand part iii of the question.

Two cars move away from a point O on a straight road at the same time, moving in the same direction.
Car A accelerates from rest with constant acceleration 5ms^-2 until it reaches a speed of 30ms^-1 which it then maintains.
Car B accelerates from rest with constant acceleration 2ms^-2.

I) Show that car A travels 90m by the time it reaches a speed of 30ms^-1 and determine the time taken

I've done this question, I've proved the displacement is 90m and I got time=6seconds


ii) Find the distance that car B has travelled in this time

I've found the answer to be 36 metres using the SUVAT eqns (as it is a constant acceleration)


iii) Car B continues to accelerate until it passes car A. At the moment that car B passes car A show that the time elapsed is approximately 27 seconds, and find the distance that both have traveled, correct to the nearest metre.

What is the question (iii) asking me to do? I don't understand! :confused:
I just need a starting point to get me going xD

(If I've gone wrong somewhere, feel free to say!)

[ebaines]

Your answers for parts I and ii are correct - so far so good.

The problem with part iii is that you have to think about the equations for car A in two distinct phases. In the first phase it's position is found from d = .5*at^2. As you found, after 6 seconds it has traveled 90 m. It's equation of motion starting at t= 6 seconds is one you would use for constant velocity, but with an initial displacement of 90 m: d = 90m + v(t-6). Now just set that equal to the equation for Car B and solve for t.

Post back and let us know if this helped.

[sleepyheadzzz]

:S Still completely lost
why is it d= 90m + v(t-6)? (the '(t-6)'? )

Tried finding the velocity for car B when time is 6... but I got 0? Which is a load of rubbish!

Is the acceleration for Car A, after 6 seconds, 0?. but that doesn't make sense either


*Aghhhh!*


-edit-
I've somehow ended up with the quadratic t^2 + 12t - 30
which solves to give t as -3.55 and -8.45

*AGHHHH*

[ebaines]

The velocity of car B at 6 seconds is found from d = at = 2m/s^2 * 6 s = 12 m/s.

The acceleration of Car A after 6 seconds is indeed 0 - it travels at constant velocity after getting up to a velocity of 30m/s.

I think what's puzzling you is that the equation for car B is
d = .5 *2m/s^2 * t^2 for all t, whereas the equation for car A changes once it stops accelerating. For t <6, its equation is:
d = .5*5m/s^2 * t^2.
But after 6 seconds its:
d =90m+ 30m/s * (t-6).

To solve for t where car A and b have the same value for d, set these equal:

.5*2*t^2 = 90+30*(t-6).

Solve for t, then use that value to determine d.

[sleepyheadzzz]

Ive got 709m (to the nearest metre)?

[ebaines]

Ive got 709m (to the nearest metre)?

Yep.