# SUVAT equation help!!

• Sep 24th 2007, 02:08 PM
nugiboy
SUVAT equation help!!
Q. Two clay pidgeons are launched vertically upwards from exactly the same spot at 1 s intervals. Each clay pidgeon has an initial speed 30 m/s and acceleration 10 m/s downwards. How high above the ground do they collide?

I know i need to find out when both the pidgeons have the same height using a suvat, but im not sure how i would use an equation as i only have the acceleration and the initial speed.

Can anyone give me a hint, or even tell me the whole method. Thanks in advanced.
• Sep 24th 2007, 08:45 PM
ticbol
SUVAT?

I know Math is a language, but suvat equations?
Wikipedia says, "The SUVAT equations are five basic equations used to describe motion of a classical system under constant acceleration. They are named SUVAT equations after the five variables that they contain."

s--- Displacement. Units of m (meters, i.e distance and direction from start. It is a vector quantity).
u--- Initial velocity. Units of ms - 1 (meters per second, i.e speed and direction. Is a vector quantity).
v--- Final velocity. Units of ms - 1 (meters per second, i.e speed and direction. Is a vector quantity).
a--- Acceleration. Units of ms - 2 (meters per second squared, i.e rate of change of speed, and direction. Is a vector quantity).
t ---Time. Units of s (seconds, i.e an amount of time. Is a scalar quantity).

v = u +at -------------(1)
s = [(u +v)/2]*t -------(2)
And 3 more formulas ---(two for s and one for v^2)---that can be easily derived from the two mentioned above.

I have memorized (1) and (2), in different forms, long time ago, and they are all I need to take care of these so-called SUVAT equations now. :)

Final velocity, Vf = Vo +at --------------------------(1')
distance travelled, d = [(Vo +Vf)/2]*t ---------------(2')

------------------------------------------

Q. Two clay pidgeons are launched vertically upwards from exactly the same spot at 1 s intervals. Each clay pidgeon has an initial speed 30 m/s and acceleration 10 m/s downwards. How high above the ground do they collide?

The two clay pigeons will collide when the first one is already going down while the second is still rising up ----at the same height above the ground.
s1 = s2 ------(i)

Given:
u1 = u2 = 30 m/sec upwards
a1 = a2 = 10 m/sec/sec downwards. ----so it is -10 m/sec/sec as we take upwards to be positive.

If we make the time the first one is fired as our reference, then
t1 = t seconds
t2 = (t-1) seconds -----because t2 is less than t1.

We use the suvat equation, s = ut +a(t^2)/2:
s1 = 30t +(-10)(t^2)/2
s2 = 30(t-1) +(-10)[(t-1)^2]/2

s1 = s2
So,
30t +(-10)(t^2)/2 = 30(t-1) +(-10)[(t-1)^2]/2
30t -5t^2 = 30t -30 -5[t^2 -2t +1]
-5t^2 = -30 -5t^2 +10t -5
0 = 10t -35
t = 35/10 = 3.5 sec

Hence,
s1 = 30(3.5) -5(3.5)^2 = 43.75 m
s2 = 30(3.5 -1) -5(3.5 -1)^2 = 43.75 m

Therefore, the two clay pigeons will collide at 43.75 meters above the ground.