# Acc Linear Motion

#### mcintyre_ie

Ok, so im having troule with yet another accelerated linear motion question... i got a 40% on my last exam, and i need at least a 90% on this one to keep a decent average... any help is desperately needed and very much appreciated :)
So here it goes:
(A) A particle is projected vertically upwards with velocityum/s and is at a height h after t1 and t2 seconds respectively. Prove that:
t1.t2 = (2h)/g

(B) A car accelerates uniformly from rest to a speed v m/s. It then continues at this constant speed for t seconds and then decelerates uniformly to rest.
The average speed for the journey is (3v)/4

(i) Draw a speed-time graph and hence, or other wise, prove that the time for the journey is 2t seconds.
(ii) If the car driver had observed the speed limit of (1/2)v, find the least time the journey would have taken, assuming the same acceleration and deceleration as in (i).

Arghhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh!

#### HallsofIvy

Please notice (did you read "Read this first", first?) that we expect people to show what they have tried when posting a problem on this forum- we won't do your homework for you!

If you gave up without trying ANYTHING, or if you honestly don't know anything about how to even start working on these, then I don't know how we can help you get a 90% on your next test. I would recommend you go to your teacher and throw yourself on his/her mercy!

Since this is "yet another accelerated linear motion question" you should know by now that v(t)= v0+ at and that x(t)= -(a/2)t2+ vo t+ x0.

(A) If "a particle is projected vertically upwards with velocityum/s and is at a height h after t1 and t2 seconds respectively", then

h= (-g/2)t12+ u t1 and
h= (-g/2)t22+ u t2.

Setting those equal should make it easy to get the result.

(B) Drawing the graph shouldn't be any problem. While accelerating at a constant rate, the graph will be a straight line, through the origin, of course, with slope equal to the accelration. The time that the car is coasting, ignoring friction, its speed is constant so the graph is a horizontal straight line. While decellerating, the graph is a straight line sloping downward. The average speed is the area under the graph (which is a trapezoid) divided by the time span.

(C) Knowing the driver's maximum speed, you know the height of the horizontal line in (B). That should let you calculate the time.

#### mcintyre_ie

Edit - The below post is what the actual problem now is.

Last edited:

#### mcintyre_ie

Ok, so for part a, you say that "setting those equal should make it easy to get the result". Basically what i took that up as meaning is " h=h " so "(-g/2)t1^2+ u t1 = (-g/2)t2^2+ u t2 " , which gives me two vairables t1 and t2, equating to each other. Where am i going wrong here? Is there some thing about the times that im missing or maybe something else pretty obvious that im missing? When i try to "fix" my answer i get " (2h)/g = 8ut - 4t^2 ". Which is very very wrong...

So that was part a, now onto part b:

Ive drawn the graph, and made a very dodgy comp reproduction:
http://thumbs.webshots.com/s/thumb4/9/81/81/91198181xFqMxj_th.jpg

All i can make out is that the area of space 1 is .5(t1)(v) = distance, area of space 2 is (t)(V) and area of space 3 is (.5)(t2)(v).
So again ive got 3 variables (t1, t2 and V). V i can keep, since the average speed is given as being = (3v)/4 (i think this is right anyway). So now that ive done some very basic (very possible that its just basically wrong too...) can anybody give me a little help ASAP?

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