Physics, 16.10.2019 03:40, Skylynn11

Aparticle is moving along a straight line such that when it is at the origin it has a velocity of 8 m/s.

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a = 4 * t^2 – 2
At t = 0, a = -2 m/s^2
At t = 2, a = 4 * 2^2 – 2 = +14 m/s^2
At t = 4.3, a = 4 * 4.3^2 – 2 = 71.96 m/s^2

The acceleration is constantly increasing. The graph of acceleration vs time a parabola. The area under this graph is the velocity. The area under the velocity graph is the displacement. We have some integrating to do.

Velocity is the integral of acceleration
Integral of 4t^2 – 2 = 4/3t^3 – 2t + c

Position is the integral of velocity
Integral of 4/3t^3 – 2t + c = ⅓ * t^4 – t^2 + c * t + d

Position = ⅓ * t^4 – t^2 + c * t + d
Initial position = 1
At t = 0, position = 1
1 = ⅓ * 0^4 – 0^2 + c * 0 + d
1 = d
Position = ⅓ * t^4 – t^2 + c * t + 1

At t = 2, position = 20
20 = ⅓ * 2^4 – 2^2 + c * 2 + 1
20 = 5⅓ – 4 + c * 2 + 1
20 = 2⅓ + c * 2
17⅔ = c * 2
8.8333 = c

Position = ⅓ * t^4 – t^2 + 8.833 * t + 1
Position = ⅓ * 4.3^4 – 4.3^2 + 8.833 * 4.3 + 1
The answer is approximately 134.45 m

answered: Guest

russian is not my first language, so i'm afraid i can't write out an answer in russian, but hopefully you can find a way to do so. here's my best translation of the problem: you're given a plot of a particle's position in one-dimensional space over time, and you're asked to find either the total distance traveled by the particle after 6 seconds had elapsed, or the total displacement of the particle after 6 seconds.

if you want the total distance traveled, you would first need to determine the particle's velocity as a function of time, then integrate that over the 6-second interval. we know the position function is quadratic with roots at 0 and 6, so we have

$x(t)=ax(x-6)$

where $a$ is unknown. because the position is parabolic, there is symmetry about the midpoint at $t=3$ , and we know that

$x(3)=10\implies -9a=10\implies a=-\dfrac{10}9$

so that the position function is

$x(t)=-\dfrac{10}9x(x-6)$

which means the velocity is

$v(t)=\dfrac{\mathrm dx}{\mathrm dt}=-\dfrac{20}9x+\dfrac{60}9$

then the total distance traveled is

$\displaystyle\int_{t=0}^{t=6}|v(t)|\,\mathrm dt=\left\{\int_{t-0}^{t=3}-\int_{t=3}^{t=6}\right\}v(t)\,\mathrm dt=20$

so the total distance traveled is 20 m.

on the other hand, if you want to find displacement: the particle starts at $x=0$ and ends at $x=0$ , so its displacement is 0.

answered: Guest

well clustered together.

points close on a scatter diagram

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Aparticle is moving along a straight line such that when it is at the origin it has a velocity of 8...