Acc Linear Motion

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).



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

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?
Thanks in advance

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