Help me with calculus qstn plz. (1 Viewer)

[redhat]

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]

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)

 

[...]

nah, it saids distance travelled...not asking for displacement...but ur right bout going back to the 'beginning'...

 

[kewpid]

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]

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

 

[Dangar]

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

 

[...]

Dangar, i am lost by ur method...

 

[TimTheTutor]

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