Physics I: Tough one-dimenional kinematics problem.

This is about about as tough as one-dimensional kinematics problems come.

The problem: A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bolt falls off the side of the rocket. The bolt hits the ground 6.0s later. What was the rocket's acceleration?

(the picture is not given, it is my drawing)

Notice that the bolt does not just fall start free falling, it has the same upward velocity as the rocket, so is more like a projectile straight upward.

The important part of this one is to pay very careful attention to what your variables are when solving this algebraically. Make sure to note what variables you know, what you don't know, and what you need to find out. In this case, we are solving for a_R.

Known:

t₀ = 0.0s

t₁ = 4.0s

t₂ = 10.0s

y_0R = 0.0m

y_0B = y_1R

v_0B = v_1R

I start by writing up some kinematic equations for both the rocket and the bolt.

y_1B – y_0B = v_0B (t₂ – t₁) – ½ g (t₂ – t₁)²

In this case, (y_1B – y_0B) is not zero, and Δt is not equal to t₂! Now what? We don’t know v_0B or y_1B, so this problem is unsolvable, right? Wrong!

Now consider the equations for the rocket:

v_1R = v_0R + a_R t₁

v_1R = a_R t₁

y_1R = y_0R + v_0R t₁ + ½ a_R t₁

y_1R = ½ a_R t₁²

Notice that not only have we solved for both of the variables in our bolt’s equation, but they are both in terms of acceleration, meaning we’ll only have 1 variable now. Back to the bolt’s equation:

y_1B – y_0B = v_0B (t₂ – t₁) – ½ g (t₂ – t₁)²

– y_0B = v_0B t₁ – ½ g (t₂ – t₁)² (Notice that v_0B = v_1R = a_R t₁, and y_0B = y_1R = ½ a_R t₁²)

– ½ a_R t₁² = a_R t₁ (t₂ – t₁) – ½ g (t₂ – t₁)²

Solve for a_R.

a_R t₁ (t₂ – t₁)  + ½ a_R t₁² = ½ g (t₂ – t₁)²

a_R (t₁ (t₂ – t₁)  + ½ t₁²) = ½ g (t₂ – t₁)²

a_R = ½ g (t₂ – t₁)²/ (t₁ (t₂ – t₁)  + ½ t₁²)

Solution: a_R = 5.5m/s²