Question

could i please have some help with this question? :)

A truck starts from rest at the top of a uniform slope and coasts down it a distance of 175m, as a result its height above the ground is reduced by 59.1m. The plane is frictionless. Whats the trucks velocity when it reaches the 175m mark and how long does it take to reach the mark?

Answer 1

um, you need a time or \[\Delta time\] to get the velocity or the velocity to get the tim

Answer 2

time*

Answer 3

yeah pretty much i just dont know how to figure that out because from whats provided theres no time just distances.

Answer 4

I know that \[V=d/t \] and \[t= d/V\] so without at least the time or the velocity, we couldn't figure out either of them (there would be two blanks, and the only thing we know is the distance)

Answer 5

yeah i know all that and thats why im stuck, and thats why i can on and asked the question :) i thought maybe there was another way of figuring it out.

Answer 6

well there might be, but I can't figure it out myself

Answer 7

okay, all good. thank you any way :D

Answer 8

I think this should help http://www.youtube.com/watch?v=9tAFhZb6DdA

Answer 9

at 3:04 is when the main part starts

Answer 10

okay thank you

Answer 11

did it help?

Answer 12

anyway, I hope it did, and if it did, no problemo :) I gotta go get some sleep so goodnight :)

Answer 13

okay thank you. have a good night.

Answer 14

velocity-time graph

Answer 15

At the top of the slope, the potential energy was mgh, and Kinetic energy 0. At the bottom Potential energy is mgh' and Kinetic energy is mv^2/2. So by the law of conservation of energy
\[mgh+0=mgh'+mv ^{2}/2\]
Take mgh' to the other side, cancel m and put h-h'=59.1.
So the velocity is
\[\sqrt{2g \times59.1}\]

Answer 16

wow that amazingly smart. but in simple terms that it

the answer is 34.0m.s^-1 but how'd you get that equation
\

Answer 17

Yeah, that is the answer.That equation is the law of conservation of mechanical energy. According to it, the total mechanical energy(Potential energy+Kinetic energy) at all times is same. Therefore, the mechanical energy of the truck when it was at top, should be the same when it was below.
Did u get it?

Answer 18

okay i get it but i dont get how you work it out, sorry.

Answer 19

like the equation.

Answer 20

Oh, for that. Just take the potential energy at a point, add it to the kinetic energy at that point. then equate this to the sum of potential energy and kinetic energy at another point.
Potential energy at top=mgh
Kinetic energy at top=0
Potential energy at bottom=mgh'
Kinetic energy at bottom=mv^2/2(U have to find v)
\[mgh+0=mgh'+mv ^{2}/2\]
\[mgh-mgh'=mv ^{2}/2\]
Cancelling m(common on both sides)
\[g(h-h')=v ^{2}/2\]
\[v ^{2}=2g(h-h') or v=\sqrt{2g(h-h')}\]
Now h-h' = 59.1m so, just susbtitute it and u will get the answer
Hope that helped

Answer 21

http://openstudy.com/study?login#/updates/4f434bcae4b065f388dc98e2

Answer 22

is there ahh nope i am completely lost

Answer 23

u could click on the link, see if it helped in anyway if your lost I'll try not to loose you, :)

Answer 24

okay ive done some raging and figuring out of my own and i come to the conclusion that the easiest way to figure this out is
V^2=u^2+2as
v^2=(0.00m.s^-1)^2+2*9.8m.s*59.1m
v^2=2*9.8m.s*59.1m
v^2=1158.36
v=sqrt(1158.36)
v=34.0m.s

Answer 25

so y ask, questions you already know the answers to ?

Answer 26

i didnt. i just worked it out.