Kinematics problem help!

**ziggy777** · August 5th 2009, 09:55 PM

OK here is a really weird looking kinematics problem ...

A body is travelling at 20 m/s when it passes point P and 40 m/s when it passes point Q. Find its speed when it is halfway from P to Q, assuming uniform acceleration.

This is what me and my teacher did-

1. A velocity - time graph with initial point at (0,20) and it ends at v=40. I called that point (t2,40). Then the point when it is halfway through is (t1,v1). I tried to equate the area of 1st trapezium with the second and called that equation 1. Then is made another equation by equating the gradients of two segmets. But at any time I only have 2 equations with 3 unknowns!

I know I'm missing a tiny thing nut can't figure it out...

thnx...

**songoku** · August 5th 2009, 10:23 PM

Hi ziggy777

I prefer to solve this problem using simple kinematic formula :
v^2 = u^2 + 2as

You can get two equations by considering :
1. when the body travels from P to Q
2. when the body travels from P to the halfway of PQ

**great_math** · August 5th 2009, 10:36 PM

Let u=20 m/s
v=40 m/s

so 40=20+at [ a is acceleration and t is total time]

so at=20 or t=20/a

Now $s=ut+\frac1{2}at^2$ [ S is the total distance]

$s=20t+\frac1{2}at\times t=20t+10t=30t$

So halfway is $s_1=\frac{s}{2}=15t$

Now let the velocity at the halfway is $v_1$

$v_1^2-u^2=2as_1=2\times a\times15t=2\times a\times \frac{15\times20}{a}$

So, $v_1^2=400+600=1000$

$v_1\approx 31.6$

**ziggy777** · August 6th 2009, 01:17 AM

thank you very much for your quick help great math and songoku! yes that is the correct answer according the book! I jast had to use some simple algebraic substitutions!

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