A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bold falls off the?

Question

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 6.0 seconds later. What was the rocket's acceleration? 
The answer I arrived at was 14.715 m/s², but I believe it is incorrect. I thought the bolt's initial velocity should equal the rocket's final velocity at the moment the bolt detaches. Here’s my reasoning for the bolt: 
Using the formula ( V_0 = -9.81 \, 	ext{m/s}^2 	imes 6 \, 	ext{s} ), I calculated ( V_f = 58.86 \, 	ext{m/s} ) to ( 0 \, 	ext{m/s} ).
For the rocket, using the equation ( V_f = V_0 + a(4s) ) with ( V_0 = 0 ) results in ( V_f = 4a ). 
Substituting one into the other gives me:
( 58.86 \, 	ext{m/s} = 4a )
Thus, 
( a = 14.715 \, 	ext{m/s}^2 ).
Could someone please help clarify my mistake? Thank you.

Anonymous · Accepted Answer

The initial speed of the bolt is not 58.86 m/s.
Let a be the acceleration of the rocket.
During the 4 sec lift off, the rocket has reached a heigth of 
h = (1/2)*a*t^2
with t=4,
h = (1/2)*a^16
h = 8*a
Its velocity at 4 sec is
v = t*a
v = 4*a
The initial velocity of the bolt is thus 4*a.
During the 6 sec fall, the bolt has the initial velocity V0=-4*a and it drops a total height of h=8*a. From the equation of motion,
h = (1/2)*g*t^2 + V0*t
Subsituting h0=8*a, t=6 and V0=-4*a into it,
8*a = (1/2)*g*36 – 4*a*6
Solving fot a,
a = 5.52 m/s^2

Elisha Runte

A rocket is launched straight up with constant acceleration. Four seconds after liftoff, a bold falls off the?

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