A small 6.90-kg rocket burns fuel that exerts a time-varying upward force on the
ID: 1275923 • Letter: A
Question
A small 6.90-kg rocket burns fuel that exerts a time-varying upward force on the rocket. This force obeys the equation F=A+Bt2. Measurements show that at t=0, the force is 139.0N , and at the end of the first 1.90s , it is 163.0N .
1-Find the constant A.
2- Find the constants B.
3-Find the net force on this rocket the instant after the fuel ignites
4-Find the acceleration of this rocket the instant after the fuel ignites.
5-Find the net force on this rocket 3.80s after fuel ignition.
6-Find the acceleration of this rocket 3.80s after fuel ignition.
7-Suppose you were using this rocket in outer space, far from all gravity. What would its acceleration be 3.80safter fuel ignition?
Explanation / Answer
F=A+Bt2
at t=0, the force is 139.0N ,
so, 139= A+B*0
so, A=139 N
and at the end of the first 1.90s , it is 163.0N .
at
F=A+Bt2
163=139+B*1.90*1.90
B= (163-139)/(1.90*1.90) = 6.64819945
net force on this rocket the instant after the fuel ignites = at t=0, F= 139 N
the net force on this rocket 3.80s after fuel ignition= F= 139+6.64819945 *3.80*3.80 =235 N
the acceleration of this rocket 3.80s after fuel ignition= 235/6.9 =34.057971 m/s^2
using this rocket in outer space, far from all gravity. its acceleration be 3.80safter fuel ignition=34.057971 +9.8 =43.857 m/s^2