Physics, 30.03.2020 21:19 iibabycarrotsii

Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t=0. For the interval 0

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Spacecrafts A and B are flying parallel to each other through space and are next to each other at time t=0. For the interval 0<t<6s, spacecraft A’s velocity vA and spacecraft B’s velocity vB as functions of t are given by the equations vA(t)=t^2−5t+20 andvB(t)=t^2+3t+10, respectively, where both velocities are in units of meters per second. At t=6s, the spacecrafts both turn off their engines and travel at a constant speed.

(a) At t=0, is the speed of spacecraft A greater than, less than, or equal to the speed of spacecraft BB?

(b) Each spacecraft has thrusters for speeding up and reverse thrusters for slowing down. At t=0, one of the spacecrafts is slowing down and one is speeding up.

i. Which spacecraft is slowing down?

ii. Calculate the time interval during which the spacecraft selected in part (b)(i) is slowing down.

(d) Is the acceleration of spacecraft A increasing, decreasing, or constant?

Explanation:

Given that

$V_A(t) =t^2-5t+20\\\\V_B(t)=t^2+3t+10$

for 0< t<6s

a) at t = 0

$V_A(0) = 20m/s\\\\V_B(0) = 10m/s$

speed of A is greter than B at t = 0

$\frac{dV_A(t)}{dt} = 2t-5\\\\\frac{dV_B(t)}{dt} = 2t+3$

The speed of A is less by 5m/s²

i) A is slowing down

ii) $\frac{dV_B(t)}{dt} =2t-5\\\\t =5/2 = 2.5sec\\$

hence, A is slowing down from 0 to 2.5sec

$V_A(t) = V_B(t)\\t^2-5t+20=t^2+3t+10\\\\10=8t\\t=1.25sec$

Hence, both are travelling at same speed at t = 1.25sec

$\frac{d^2V_B(t)}{dt^2} =\frac{d(2t-5)}{dt} =20$

Hence the acceleration of space craft A is increasing

Answer from: Quest

The next element is345640

Answer from: Quest

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.

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