Question

A snail moving across the lawn for her evening constitutional crawl is attracted to a live wire.
on reaching the wire her speed increases at a constant rate&it doubles from 0.001 ms^-1 in 10sec.
she remains at this speed for a further 15sec. while she remains contact with the wire.
how to calculate the distance traveled during this time???
thanks :)

Answer 1

Double speed = .002ms^-1
Average rate of change = .002+.001=.003 divided by 2 = .0015.
.0015 x 10 = .015
.002 x 5 remaining seconds = .01
.01+.015 = .0225 ms^-1

Answer 2

constant acceleration equation
Distance traveled=(Initial Velocity)*(time) + (acceleration)*((time)^2)
find the average acceleration and use that formula over the first 10 secs

for the next 15 seconds u can use distance=(velocity)*(time)

Answer 3

ive never seen this formula before,
I DONT KNOW HOW TO USE IT *panic*

Answer 4

which one?

Answer 5

the distance travelled one :S
im not even sure whether ive got the number that id need for it :(

Answer 6

But the answers are using two different formula.

Answer 7

Maybe take a look here: http://en.wikipedia.org/wiki/Acceleration

Answer 8

r u in calculus?

Answer 9

i dunno, it mechanics 1 motion :S

Answer 10

have u done integration b4?

Answer 11

thats dy/dx stuff right ??

Answer 12

yes.

Answer 13

ok basically to use the equation i gave u need an acceleration n it needs to b a constant. u multiply this acceleration by the time u r accelerating squared so \[t^2\]

Answer 14

then u add this to ur initial velocity(the velocity u started out at) multiplied by time so the whole equation is \[d=vt+at^2\]

Answer 15

v is the velocity u started out at in this case its .001ms^-1

a is the acceleration u have to find. since u know the final acceleration which is .002ms^-1 since it doubled, u can find the average acceleration. (final velocity - initial velocity)/time so:
(.002-.001)/10=
.001/10=
.0001ms^-2

t is the time u travaled in this case its 10 seconds

plug all of the numbers in and u will find the distance for the 1st part

Answer 16

Isn't Jakesmilyface formula equivalent to yours?

Answer 17

u know, the (u+v)/2

Answer 18

yea, but he explained and he didnt :)

Answer 19

for the 2nd part u can use the same exact equation but the acceleration this time is 0 because it stayed at the same speed thus no acceleration

so this time the velocity is .002ms^-1
the acceleration is 0 and
the time is now 15 seconds

plug in all the numbers n u can get the answer. Then add it to ur first answer n u will get the answer u need

Answer 20

yes it is i think

Answer 21

I agree, it's a great explanation.

I'm just asking if he agrees that the 2 are equivalent.

Answer 22

???
second answer :S
im confused

Answer 23

but i did what you said and came up with 0.003
but then i didnt use the thing that was just being worked out the 0.0001ms^-2

Answer 24

its a 2 part problem because at 1point the snail accelerated which means u have to find the distance he traveled while he was accelerating and then u have to find the distance he traveled at his top speed.

Think of it as when u r driving a car on the highway when ur accelerating to the speed limit(hopefully no higher) u r still traveling some distance right? u need an equation to solve the amount of distance u traveled then n when u get to ur top speed u need to use a different equation because u r no longer accelerating

Answer 25

but doesnt that mean that i should have used that 0.0001 to work out the answer??
cause i didnt and got 0.03, is that right??

Answer 26

I still like to know if you agree that the formulas for part 1 are equivalent?

Answer 27

the answer should b .05m

Answer 28

oh :(
WHY?!?!
why?!?!
why cant I do it :s

Answer 29

ok im gonna work it out n show my work

Answer 30

thanks you :)

Answer 31

1st part
d=vt+at^2
v=.001
a=(.002-.001)/10=(.001)/10=.0001
t=10
so it when i replace everythin it becomes
d=(.001)(10)+(.0001)(10^2)=
.01+.01=.02

2nd part
d=vt+at^2
v=.002
t=15
a=0
so it becomes
d=(.002)(15)=
.03

add the 1st n 2nd part together
.02+.03=.05
so in that 25 seconds it traveled .05m

Answer 32

also i looked at jakesmileyface's work n it isnt correct

Answer 33

accelerating from one speed to another is not the same as driving at the average of the speeds

Answer 34

thanks your working out really helped me :)

Answer 35

anytime