a particle moves along a line with velocity function v(t)=t^2 - t...?

Question

a particle moves along a line with velocity function v(t)=t^2 - t, where v is measured in meters per second. Find displacement and find: the distance traveled by the particle during the time interval [o,5]

i need some help with this problem, i appreciate anyone who tries to answer and explain it ^^

Answer 1

The derrivative of a displacement equation is its velocity equation.
Therefore, you have to take the integral of your given velocity equation to come up with the displacement function.

The integral of v(t) = t^2 - t is...

d(t) = (t^3)/3 - (t^2)/2 + c

when t = 0, c = 0 assuming that the particle starts from a reference point zero.

Therefore, d(t) = (t/3)^3 - (t/2)^2

When...
t = 0, d(t) = 0
t = 1, d(t) = -0.1666
t = 2, d(t) = 0.6666
t = 3, d(t) = 4.5
t = 4, d(t) = 13.333
t = 5, d(t) = 29.166

The total distance traveled by the particle is the sum of all values for d(t) which is...

47.5 meters

The total displacement after 5 secs is the value of d(t) for t =5 which is...

29.166667 meters

If you do get the integral of v(t) and evaluate if from 0 to 5 you will get 29.16667 meters and not 112.5 meters!!!

Answer 2

the displacement is just the how far it traveled and you just take the integral from 0 to 5 of v

int (t^2-t) dt = t^3/3 + t^2/2
then plug in 5 and subtract with 0 plugged in (i dont have my calculator)

the difference between distance traveled and displacement is can be seen like this; if the particle went back and forth for a while and ended up at the same place the displacement would be 0 and the distance traveled would be the distance traveled.

So how you find it from the velocity function is find out where v is negative (moving backwards) and break the integral up into peices and put a negative sign on the parts in which v is negative. (look at the graph)

so this v is negative between 0 and 1and positive from 1 to 5 so the distance traveled is

- integral from 0 to 1 of v dt + integral from 1 to 5 of v dt

(notice the negative sign on the first integral)

hope that helps

Answer 3

Velocity function is the derivative of the position function, so to find the displacement at a particular point you need to find the integral of the velocity function, here x(t)=1/3t^3-0.5t^2(position), displacement at a given time is found by finding the area under the velocity graph or evaluating the position function at x(0)=0,
x(5)=112.5, so displacement between 0 and 5 sec is 112.5. Hope this helps.

Answer 4

To obtain the displacement, you have to integrate.

Remember these simple rules to that:

1) You have to integrate every term

2) Antiderivative from t^n is t^(n+1)/(n+1)

Example:

From t, you obtain t^2/t
From t^2, you obtain t^3 / 3
From t^3, you obtain t^4 / 4
etc.

From 1 you obtain t

3) If you have a number, ie 5 t^2, you obtain 5 t^3/3.

4) And you always have to add a constant, which could be evaluated in certains cases, and its not needed in most of them

Now your exercise:

This is what you obtain

d(t) = t^3/3 - t^2/2 + C

Then you evaluate this in 0 and in 5:

d(5) = 5^3/3 - t^2/2 + C
d(0) = C

Since you you have to substract, d(5)-d(0), you dont need to calculate C.

Ana

Answer 5

Displacement is the crucial of speed with appreciate to time. So: x(t) = crucial of v(t) x(t) = -(a million/3)t^3 -3t^2 +6t +C the position C is a continuing. => x(-a million) = -(a million/3)(-a million)^3 -3(-a million)^2 +6(-a million) +C = (a million/3) -3 -6 +C = -26/3 +C and x(5) = -(a million/3)(5)^3 -3(5)^2 +6(5) +C = -100 twenty 5/3 -seventy 5 +30 +C = -10/3 +C subsequently the adaptation is -10/3 +C - (-26/3 +C) = -10/3 +C +26/3 -C = 16/3 = 5.333...