Solving Complex Math Problems: Deceleration, Acceleration, & Earthquakes

[BriannaUND]

i don't know why I'm having so many problems but I greatly appreciate the help!
1) A train normally travels at a uniform speed of 72 km/hr on a long stretch of straight, level track. On a particular day, the train must make a 2.0-min stop at a station along this track. If the train decelerates at a uniform rate of 1.0m/s2 and, after the stop, accelerates at a rate of .5 m/s2, how much time is lost because of stopping at the station?
So far I have calculated vo= 72 km/hr (.278 m/s divided by 1 km/hr) = 20.0 m/s. Then t= v-vo/a = 0-20 m/s divided by -1.0 m/s2 = 20 sec to stop. Does this look correct? And how do I go about calculating the time lost?
2)The driver of a pickup truck going 100 km/hr applies the brakes, giving the truck a uniform deceleration of 6.5 m/s2 while it travels 20.0 m. a) what is the speed of the truck in km/hr at the end of this distance? and b) how much time has elapsed?
3) An earthquake releases two types of traveling waves, called transverse and longitudinal. The average speeds of transverse and longitudinal seismic waves in rock are 8.9 km/s and 5.1 km/s respectively. A seismograph records the arrival of the transverse waves 73 s before that of the longitudinal waves. Assuming that the waves travel in straight lines, how far away is the center of the earthquake from the seismograph?
Thanks again!

 

[quasar987]

You don't seem to have put much effort into this.

So far, 1) is perfect. Think, and keep going.

2) & 3) Yeah, so what are your thoughts on how to tackle these problems?


P.S. Please post further threads in the homework section: https://www.physicsforums.com/forumdisplay.php?f=15. Thanks.

 

[BriannaUND]

As for
#2a) I've calculated v2 = 100 km/hr + 2 (-6.5 m/s2)(20 m), v2 = 360, v= 18.9, 100-18.9 = 81.1 km/hr
#2b) vx= vox + axt making 81.1 = 100 + (-6.5)t making t= 2.9 sec
#3) I've tried calculating average velocity and then plugging that answer in for x= average velocity x time for both of the waves. But I'm getting confused because the problem states that the arrival of the t-waves "73 sec before l-waves" so I'm not sure what to plug in for t when calculating the distance of the l-waves

 

[VietDao29]

#2a is wrong. So #2b is also wrong. You are on the right track but your calculation is wrong.
v_f ^ 2 = v_i ^ 2 + 2ad. You should also convert the 100 km / h to m / s.
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The transverse waves is faster than the longitudinal waves, therefore the transverse waves will reach the seismograph before the longitudinal waves (73 sec before the l-waves).
The two types of waves are released at the same time. If it takes the t-waves t (seconds) to reach the seismograph, then it takes the l-waves t + 73 seconds to reach the seismograph. So you have the time each kind of waves need to reach the seismograph. To reach the seismograph, they travels the same distance. So:
v_l (t + 73) = v_t t
You know the v_l : speed of l-waves, v_t : speed of t-waves. You can find the amount of time (t) it takes the t-waves to reach the seismograph. From there, you can easily find the distance from the seismograph to the center of the earthquake.
Viet Dao,