- #1

mcintyre_ie

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Ok, so I am having troule with yet another accelerated linear motion question...

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).

Ok, so for part a, I've been told 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 variables t1 and t2, equating to each other. Where am i going wrong here? Is there some thing about the times that I am missing or maybe something else pretty obvious that I am 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:

Graph Pic

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 I've 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 I've done some very basic (very possible that its just basically wrong too...) stuff, can anybody give me a little more help ASAP?

Thanks in advance