Answer
a. $2$ sec, $190$ ft/s
b. $2$ sec.
c. $8$ sec. $0$ ft/s.
d. $10.8$ sec, $90$ ft/s (downward).
e. $2.8$ sec.
f. $2$ sec.
g. $2\leq t\leq 10.8$ sec. $-32 ft/s^2$
Work Step by Step
a. When the engine stopped, the rocket lost power and its speed will start to decrease; from the given figure, we can identify this point at about $t=2$ sec, which gives a velocity value of about $190$ ft/s
b. From the given figure, we can identify that the engine burned for about $2$ sec.
c. When the rocket reaches its highest point, its velocity will change sign (from going upward to downward) and we can identify this point at about $t=8$ sec. Its velocity then can be found as $0$ ft/s.
d. When the parachute pops out, its velocity will suffer a sudden change and we can identify this point at about $t=10.8$ sec. At this time, the rocket was falling at a speed about $90$ ft/s (downward).
e. The rocket started falling at about $t=8$ and the parachute opened at about $t=10.8$; thus the time gap is about $10.8-8=2.8$ sec.
f. The rocket’s acceleration reached its greatest value when the derivative of the velocity is the biggest, which can be identified from the graph as about $t=2$ sec.
g. The acceleration would be a constant when the slope of the velocity curve was a constant, which happened in the interval of $2\leq t\leq 10.8$ sec. From a calculated slope value, we can find this acceleration to be about $a=\frac{-90-190}{10.8-2}\approx-32 ft/s^2$
