Find The Total Displacement And Total Distance Traveled Over The Time Interval [0 9].

Find the total displacement and total distance traveled over the time interval [0,9].?

A particle moves in a straight line with velocity 10 - 2t ft/s. Find the total displacement and total distance traveled over the time interval [0,9].

what is Displacement (ft)?
and
what is Distance (ft)?

Thank you

Nd the total displacement and total distance traveled over the time interval [2,5]?

If it moves x feet in t sec velocity v=ds/dt=t^-2 -1/16 =(16-t^2)/(16t^2)
and v>=0 for t^2<=16 or t<=4, and it then moves to the right.
v<0 for t>4 and it then moves to the left.
a) Integrating gives x=-1/t-t/16+c and if x=0 when t=2 c=5/8
so x=-1/t -t/16+5/8 and when t=5 x=?? which is the total displacement .
b) v=0 when t=4 so find the displacement to t=4 which gives x=1/8
When t=5, x=9/80 so particle starts from x=0 and moves to x=1/8 (=10/80)
then moves back to x=9/80 by distance 10/80-9/80=1/80
so total distance =10/80 +1/80=11/80 feet.

Displacement and total distance traveled....?

Note that, given a velocity function of v(t), the total displacement on [a, b] is given by:
∫ v(t) dt (from t=a to b).

Then, the total distance traveled on [a, b] is:
∫ |v(t)| dt (from t=a to b).
(Note that the only difference is the absolute value signs)

So, we see that the total displacement is:
∫ (32 - 2t^2) dt (from t=0 to 9)
= [32t - (2/3)t^3] (evaluated from t=0 to 9)
= (288 - 486) - (0 - 0)
= -198 ft.
(Note that displacement can be negative!)

The total distance traveled is:
∫ |32 - 2t^2| dt (from t=0 to 9).

By splitting up the integral at t = 4:
(Note that 32 - 2t^2 > 0 for 0 < t < 4 and 32 - 2t^2 < 0 for 4 < t < 9)
∫ |32 - 2t^2| dt (from t=0 to 9)
= ∫ |32 - 2t^2| dt (from t=0 to 4) + ∫ |32 - 2t^2| dt (from t=4 to 9)
= ∫ (32 - 2t^2) dt (from t=0 to 4) - ∫ (32 - 2t^2) dt (from t=4 to 9)
= 256/3 - (-850/3)
= 1106/3 ft.

I hope this helps!

Can you help with this displacement and total distance traveled integral problem?

Yes, that's what it means. It means that the particle is oscillating back and forth. Sometimes the velocity is positive (moving in the +x direction), sometimes it is negative (moving in the -x direction), and at the points where the direction changes, the velocity is 0.

The difference between displacement and distance is that if the particle moved from 0 to +10 and back to 0 for instance, the displacement is 0 (it's back where it started) but the distance is 20. It moved 10 units forward and another 10 units back. So you need to separate the segments where v is positive from the segments where v is negative and figure the distance separately for each.

Find the displacement and the distance traveled over the indicated time interval?

Displacement is simply the distance between starting point and ending point
r(0) = (1-3sin(0), 3cos(0)) = (1-3(0), 3(1)) = (1,3)
r(3π/2) = (1-3sin(3π/2), 3cos(3π/2)) = (1-3(-1), 3(0)) = (4,0)

Displacement = √((4-1)²+(0-3)²) = √18 = 3√2

--------------------

Total distance traveled will be length of curve from t = 0 to t = 3π/2

x(t) = 1 - 3 sin(t) ------> dx/dt = -3 cos(t)
y(t) = 3 cos(t) ---------> dy/dt = -3 sin(t)

L = ∫[0 to 3π/2] √((dx/dt)² + (dy/dt)²) dt
L = ∫[0 to 3π/2] √(9 cos²t + 9 sin²t) dt
L = ∫[0 to 3π/2] √9 dt
L = ∫[0 to 3π/2] 3 dt
L = 3t |[0 to 3π/2]
L = 9π/2

Ματπmφm

Find the distance traveled over given time interval.?

The integral of velocity is displacement. Your expression is correct. For the displacement from t = 0 to t = 3, that gives [27/3 + 9 - 9] - 0 = 9.

However, I think the question is asking for something different. This velocity factors as (t + 3)(t - 1). It is negative up till t = 1, after which it is positive. In other words, the object moves in the positive direction from t = 0 to t = 1, and then the other way from t = 1 to t = 3. I think you're being asked for the total distance it travels, which means you have to separately add the two pieces.

That's why I was careful to say "displacement" in my first sentence, which is the distance from start point to end point, a different concept.

Using the same formula, the position at t = 1 is 1/3 + 1 - 3 = -5/3. The object moves from 0 at t = 0 to -5/3 at t = 1, a distance of 5/3. Then it moves from -5/3 at t = 1 to +9 at t = 3, a distance of 32/3. The total distance traveled is therefore 5/3 + 32/3 = 37/3.

Displacement and Total Distance Traveled? - Calculus help? 10 pnt BA?

6) The velocity of an object on the interval 0<= t <=10 is shown in the graph below. If v(t) is measured in feet per second, find the object’s displacement on the interval. (use the graph I attached)


7) Refer to the graph. Find the object’s total distance traveled on the interval
0<=t <=10.

I know that displacement is end position subtract the initial position. The problem is I don't understand what the graph is showing me. Please help, and please explain how you got your answer. I want to be able to do this one my own!



Bonus help:

Use the following information for Problems 11-15:

From its initial position of s(0) = -3, a particle moves along a linear path with a known velocity of
v(t) = t^2 – 3t – 4, where v(t) is measured in meters per second and t>=0.

11.) Find the time when the particle has an acceleration of 9 meters per squared second.

12) Find a function to represent the position of the particle at any time t.

13. Find the position of the particle when its velocity is zero.

14) Determine when the velocity of the particle is positive and when it is negative.

15) Find the distance traveled by the particle on the interval 0<= t<= 5.

Find the total distance that the particle travels over the given interval?

v(t) = (t-8)(t+7)
So the particle changes direction after 8 seconds.

a) Integrate v = t² - t - 56 to get displacement, d = ⅓t³ - ½t² - 56t + k.
As no initial conditions are given we cannot determine the value of the constant, k.

b) When t = 1, d = 1/3 - 1/2 - 56 + k = k - 337/6
When t = 8, d = 512/3 - 32 - 448 + k = k - 928/3
When t = 9, d= 243 - 81/2 - 504 + k = k - 603/2
Distance travelled between t=0 and t=8 is 928/3 - 337/6 = 1519/6
Distance travelled between t=8 and t=9 is 928/3 - 603/2 = 47/6
Total distance travelled = 1519/6 + 47/6 = 1566/6.
That's approximately 261 feet.