# Accelerated Linear Motion

Hey
Id really appreciate any help anybody can give me with this problem, its urgently needed and very very important, ive been stuck on it for about an hour and dont even know where to start:
Two cars A and B, each 5m in length, travel with constant velocity 20 m/s along a straight level road. The front of car A is 15m directly behind the rear of car B. Immediately on reaching a point P, each car decelerates at 4m/s^2.
1) Show that A collides with B
2) At what distance from P does the collision occur?
3) Show the motion of both cars on the same speed-time graph.

## Answers and Replies

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FZ+
What you need to do first is to translate the "when they reach P" business to a time difference. Indeed, this problem is probably easier if you do part 3 first.

From the data given, it is clear that the two cars are 20m apart, considering their middles or both their fronts etc. Therefore, when car B passes P, car A will be 20m from it. Using the velocity it gave you, you can then find the time difference between the two reactions, and you can easily draw a speed time graph from this.

So, you know that the area under the speed time graph is distance, right? In this case, you can consider the area for B as a trapezium (Using equation area = 0.5 * h * (a+b)), and for A as a rectangle and a trapezium. Now, just locate situation at which the area for A is exactly 5 less than the area for B (Since we are considering the middles of cars in the graph, not the relative distance from front and rear) Then that's 1 and 2 sorted.

If you wanted though, you could have split the situation up as before and applied formulae like s = ut + 0.5 at^2... but that's probably overcomplicating.

FZ+ You cant imagine how grateful i am for your help... ive been trying to do this one question for like an hour, but couldnt figure out what i was supposed to do with the info given, its very different to our normal exam questions.

Using the velocity it gave you, you can then find the time difference between the two reactions, and you can easily draw a speed time graph from this."
Which reactions would i be finding the time-difference for?

I am unfamiliar with the "trapezium" things, i guess ive never met the technical term for whatever it is,

If you wanted though, you could have split the situation up as before and applied formulae like s = ut + 0.5 at^2... but that's probably overcomplicating
The laws are the way we usually try and attept it, and to be honest, im still a bit lost, so if you could just give me over-tired brain another few hints i would be sooooooooooooooo grateful
Thanks for the help so far too by the way!

FZ+
Which reactions would i be finding the time-difference for?
The time at which car B starts deaccelerating, and car A starts deaccelerating...

I am unfamiliar with the "trapezium" things, i guess ive never met the technical term for whatever it is,
Imagine a rectangle, but with the lengths of the sides different. ie. a quadrilateral (four sided shape) with two parallel sides...

Horrific ascii art follows...

Code:
| \
|   \
|     \
|       \
|         \
|          |
|          |
|          |
----------

is a trapezium.
The laws are the way we usually try and attept it
Funnily enough though, the equations you usually get given are derived effectively working out the area, but a mathematical method known as calculus.

But no matter...

If you look at the motion of car B, you can see that it follows an equation for uniform decceleration,

s = ut + 0.5 a t ^ 2

Meanwhile, the other car undergoes a bit of constant velocity first, and then switches to uniform acceleration. The other car begins with a 15 metre lead between the front and rear, but also a 20/u second head start. But u is also reduced by v = u + at

So, at time t (starting from when the second car A crosses the point), the distance car A is behind car B is given by distance = (u + a)(t + 1) + 0.5 a(t + 1)^2 - u(t) - 0.5 a(t) ^2 + 15.

When distance = 0, then there is the collision. Feed the calculated t back into the mix...

OK... I probably confused myself back there, so check my method...