# Simple Acceleration Problem throwing me off

I am having trouble with this problem from my homework. I have gone over it many times, and I must have some concept wrong.

A rocket is launched from the ground with constant acceleration. After 4 seconds, a bolt falls off and hits the ground 6 seconds later. Gravity is -9.8m/s^2. What was the rocket's acceleration?

So here's my thoughts.

## First, let's get the height of the rocket when the bolt falls.

Fell at 9.8m/s^2 for 6s, so the final velocity was 58.8m/s

Average velocity was half of that, because of the constant acceleration, so 29.4m/s

Total distance is that average, times 6 seconds, 176.4m.

Good so far?

## Next, the rocket's acceleration

We know it was 176.4m in the air after 4 seconds.

So the average velocity 44.1 m/s, final velocity is 88.2 m/s

It got to that in 4 seconds, so the acceleration must be 22.05 m/s^2

But it's not. The correct answer is 5.5 m/s^2, a quarter of what I got.

## So what went wrong?

I've tried looking through all the steps a bunch of times, and I can only think of one step that doesn't seem quite right. The bolt falling off the rocket makes it sound like it just drops, but maybe it starts that drop with an upwards velocity?

But I can't see any way of getting to that velocity. Without knowing how high the rocket is, we can't know how fast it's going, so we can't know how fast the bolt was moving upwards. And then we have two variables in the equation, and an infinite number of correct pairs.

I must be missing some concept, some formula must be missing. Can you find it?

• "Fell at 9.8m/s^2 for 6s, so the final velocity was 58.8m/s" That's the magnitude of the change in speed from the time the bolt fell off the rocket to the time it hit the ground. – Alfred Centauri Sep 17 '13 at 1:14
• Alfred, thanks. I hadn't thought if it like that. But I don't see a link between that and a different answer? How could I find the initial velocity of the bolt? – fnsjdnfksjdb Sep 17 '13 at 1:37
• The initial velocity of the bolt is zero. Then, there is constant acceleration upward for 4 seconds followed by constant acceleration downwards for 6 seconds. Think about it, what must be equal for both periods? – Alfred Centauri Sep 17 '13 at 1:41
• The distance travelled? – fnsjdnfksjdb Sep 17 '13 at 1:52

Since the initial speed an position are zero, the position and upward velocity after 4 seconds are:

$z(4) = \dfrac{a_1 (4s)^2}{2}$

$v(4) = a_1 (4s)$

When the bolt falls off, the acceleration changes from $a_1$ to $-9.8\dfrac{m}{s^2}$ and the bolt impacts the ground 6 seconds later. So...

$z(10) = 0 = z(4) + v(4)\cdot 6s - 9.8\dfrac{(6s)^2}{2} = \dfrac{a_1 (4s)^2}{2} + a_1 (4s)\cdot6s - 9.8\dfrac{(6s)^2}{2}$

Solve for $a_1$.