• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

Help me with calculus qstn plz. (1 Viewer)

redhat

Member
Joined
May 7, 2003
Messages
33
The position x m at time t s (t is greater than or equal to 0) of a moving particle moving in a straight line is given by x = t^2 - 5t +4

What is the distance travelled in the first 4 seconds?
 

Huy

Active Member
Joined
Dec 20, 2002
Messages
5,240
Gender
Undisclosed
HSC
N/A
wouldn't it be 0?

total displacement = 0 since it goes back to the beginning?

when t = 4
x = 4^2 - 5(4) +4
= 0 m

x' = velocity (or dy/dx)
x'' = acceleration (or d2y/dx2)

doh, i'm still in the english mode
plus i suck at maths hehe :)

when t=1, x = 0
when t=2, x = -2
when t=3, x = -2
when t=4, x=0

0+(-2)+(-2)+0
= -4 m? that doesn't seem right (4m to the left of where you began)

yes, i really suck (but i can do annuities/time payments/supperannuation because the sequences are cool to derive) :)
 

...

^___^
Joined
May 21, 2003
Messages
7,723
Location
somewhere inside E6A
Gender
Male
HSC
1998
nah, it saids distance travelled...not asking for displacement...but ur right bout going back to the 'beginning'...
 

kewpid

Member
Joined
Jun 1, 2003
Messages
51
Location
Sydney
Gender
Male
HSC
2003
first sketch the graph

then you can see that:
from 0 -> 1 it travels 4m
from 1 -> 2 it travels 2m
from 2 -> 2.5 it travels 0.25m
from 2.5 -> 3 it travels 0.25 m
from 3 ->4 it travels 2m

add em up and it comes to 8.5m

im sure theres a prettier way of doing it. but this still works :)
 

wogboy

Terminator
Joined
Sep 2, 2002
Messages
653
Location
Sydney
Gender
Male
HSC
2002
The best way to do this sort of question is to draw a nice clear graph of x vs t. On this graph, put t where you would normally put x (left-right direction), and put x where you would normally put y (up-down direction).

You'll find that this graph is a parabola concaving up (i.e. smiley shape), and by factorising you find that it has roots at t=1 & t=4. You can find it's vertex to be at t=2.5 (the average of 1 & 4), and therefore x= -2.25. This is the minumum point of the parabola (i.e. the furthest down it travels).

-from t=0 to t=2.5 it has travelled 6.25m (going from x=4 to x=-2.25)

-from t=2.5 to t=4 it has travelled 2.25m (going from x=-2.25 to x=0)

{note that I've split it up where the minumum pt of the parabola is i.e. at t=2.5 You don't need to split it up any further}

so the total DISTANCE travelled = 6.25+2.25 = 8.5m
 
Last edited:

Dangar

Member
Joined
Apr 18, 2003
Messages
125
Location
Sydney
you can also do it by differentiating to get an expression for velocity, and then if you solve this you get the time when the particle is at the endpoints of the motion. i.e. where velocity equals zero and so the particle turns around and heads in the opposite direction. Then you sub these values for t into the original displacement equation and add your answer= you gett the total distance travelled.
A graph is probably easier, but sometimes i just get blocks, and im not very good at graphs
 

TimTheTutor

HereToHelp
Joined
Mar 11, 2003
Messages
8
Gender
Male
HSC
N/A
x = t2 - 5t - 4 (t>=0)
v = 2t - 5
a = 2
Change in direction will only occur when v=0 and a<>0
0 = 2t - 5
t = 2.5 => x = -2.25
When t=0 x=4 and when t=4 x=0
[--->----]
[---<---------------<----]
|---------0--------------4--------------------->x
-2.25

Distance travelled = total length inside [] = 6.25 + 2.25 =8.5m
 
Last edited:

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top