Solve Kinematic Problems: Tips & Advice

[HWGXX7]

Hello,

I'am doing some selfstudy to aqcuire more experience with mechanics. I got a geat book I'am currently reading. But got stuck in a question I can't answer...


I hope you guys can get me on the good track, just need an approach to tackle te problem. Did already solve problems,but this one seems odd to me...So I wait for good tips. Thank you.

 

[jehan60188]

you have an initial velocity/position 200*10^3, 600
you know that deceleration is constant
you have a final velocity/position 50e3, 586

find the time it takes to go 14 feet with constant deceleration from 200e3 to 50e3

then, at 30 meters, he decelerates again, to zero m/s
so find that time, and sum them up

 

[HWGXX7]

a=constant, v(t)=-a.t+v_{0}, x(t)=-a\frac{t^{2}}{2}+v_{0}.t+x_{0}
Initial conditions:
v_{0}=55.56 m/s
x_{0}=0 m
Situation after t_{1} sec.:
v(t_{1})=13.89 m/s
x_(t_{1})=14 m

Rewrite equations:
-a.t_{1}=13.89-55.56=-41.67 m/s
14=(-41.67).\frac{t_{1}}{2}+55.56.t_{1}=(-20.84+55.56).t_{1}

Solution:
t_{1}=0.4032 sec.
This answer is the input to the velocity equation to get the constant decceleration, which is 103.3m/s^{2}

Travel time to reach the 30m position: \frac{556m}{13.89 m/s} = 40.03 s

Total travel time so far: 40.43 s
The decceleration time tot reach ground level can be get with the same procedure.
Total travel time: 44.73 s

Thank for the good help, relative easy problem actually.