Time & Speed- Need to check answers

[reyes117]

A locomotive is accelerating at 1.6m/s^2.It passes through a 20m wide crossing in a time of 2.4s . After the locomotive leaves the crossing , how much time is required until its speed reaches 32m/s?



=== Im using the equation x= Vot + 1/2 at^2

This is what i did, but im not sure about it. I just want to make sure my procedures are correct. I don't know if i moved the question around correctly.


x = Vot + (1/2)at^2
20 = Vo(2.4) + (1/2)(1.6)(2.4)^2
Vo = [20 - (1/2)(1.6)(2.4)^2]/2.4
Vo = 6.41 m/s, speed of locomotive when leaves the
crossing

t = (Vf - Vo)/a
t = (32 - 6.41)/1.6
t = 15.99 or 16 sec



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I need help on this question as well. Some one told me to use this equation mgh = (1/2)mV^2.......but i have never seen it before and my teacher didn't mention it. Is there any other equation i can use? the ones my teacher has gone over are V= Vo + at, x+ 1/2(Vo + V)t, x= Vot + 1/2at^2, V^2+ Vo^2 + 2ax



The greatest height reported for a jump into an airbag is 99.4 m by stuntman Dan Koto. In 1948 he jumped from rest from the top of the Vegas World Hotel and Casino. He struck the airbag at a speed of 39 m/s (88 mi/h). To assess the effects of air resistance, determine how fast he would have been traveling on impact had air resistance been absent.



***** Any help would be deeply appreciated! Thanks in advance!****

[Vanagib]

First question seems just fine to me

Second:

mgh = 1/2 mv^2 is using an energy argument instead of suvat equations to approach the problem

mgh = gravitational potential energy
1/2mv^2 = kinetic energy

So you assume that all potential energy is converted to kinetic - unfortunately as you do not know his mass you cannot use an energy argument. Instead you must use suvat.

EDIT: Actually I'm wrong here, stupid oversight.

if mgh = 1/2mv^2
2mgh = mv^2
2gh = v^2
v = sqrt(2gh)

Using this solution v = sqrt (2*9.8*99.4)
= sqrt (1 948.24)

= 44.1388...
so = 44.1ms^-1


s=99.4
a= 9.8
u = 0

v^2 = u^2 + 2as
v^2 = 0 + 2 (9.8*99.4)
v^2 = 1948.24
v = 44.1388...

so v = 44.1 ms^-1

[method_man]

A locomotive is accelerating at 1.6m/s^2.It passes through a 20m wide crossing in a time of 2.4s . After the locomotive leaves the crossing , how much time is required until its speed reaches 32m/s?



=== Im using the equation x= Vot + 1/2 at^2

This is what i did, but im not sure about it. I just want to make sure my procedures are correct. I don't know if i moved the question around correctly.


x = Vot + (1/2)at^2
20 = Vo(2.4) + (1/2)(1.6)(2.4)^2
Vo = [20 - (1/2)(1.6)(2.4)^2]/2.4
Vo = 6.41 m/s, speed of locomotive when leaves the
crossing Untill now, you did o.k. Expect for the fact that the speed vo is the speed that locomotive has at the start of the crossing. Speed at the end of the crossing (vc) would be vc=vo+at=10,25 m/s. So in your next equation instead of vo, you have to put vc.
t=(vf-vc)/2


The greatest height reported for a jump into an airbag is 99.4 m by stuntman Dan Koto. In 1948 he jumped from rest from the top of the Vegas World Hotel and Casino. He struck the airbag at a speed of 39 m/s (88 mi/h). To assess the effects of air resistance, determine how fast he would have been traveling on impact had air resistance been absent.
You can use the connection between kinetic and potential energy.
P.O.=K.O.
Ep1-Ep2=Ek2-Ek1
mgh1-mgh2=1/2*mv22-1/2*mv12
h2=0 and v1=0
mgh1=1/2*mv22
v=sqrt2gh