# Kinematic problem

1. Mar 18, 2012

### 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.

2. Mar 18, 2012

### 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

3. Mar 19, 2012

### 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, wich 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.