Physics, 10.07.2019 19:30 Sabby5069

The driver of a pickup truck going 89.9 km/h applies the brakes giving the truck a uniform deceleration of 5.99 m/s^2 while it travels 19.3 m. a. the velocity of the truck in m/s at the end of this displacement is m/s. b. how much time has elapsed?

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Answer from: juiceyj9811

(a) First, convert the truck's velocity from km/h to m/s:

$89.9\,\dfrac{\mathrm{km}}{\mathrm h}\cdot\dfrac{10^3\,\mathrm m}{\mathrm{km}}\cdot\dfrac{\mathrm h}{3600\,\mathrm s}=25.0\,\dfrac{\mathrm m}{\mathrm s}$

Then the velocity after traveling $\Delta x=19.3\,\mathrm m$ satisfies

$v^2-{v_0}^2=2a\Delta x$

where v_0 and are the truck's initial velocity and acceleration, respectively. So we have

$v^2-\left(25.0\,\dfrac{\mathrm m}{\mathrm s}\right)^2=2\left(-5.99\,\dfrac{\mathrm m}{\mathrm s^2}\right)(19.3\,\mathrm m)$

$\implies v^2=394\,\dfrac{\mathrm m^2}{\mathrm s^2}$

$\implies v=\pm19.8\,\dfrac{\mathrm m}{\mathrm s}$

Taking the square root gives two possible velocities of the same magnitude but in opposite directions. It's technically feasible that over the given displacement, the truck changes its direction and is accelerating back towards its starting position since we're treating the truck as a particle.

(b) When acceleration is constant, average velocity is given by

$v_{\mathrm{av}}=\dfrac{x-x_0}t=\dfrac{v+v_0}2$

which tells us that

$t=\dfrac{2(19.3\,\mathrm m)}{v+25.0\,\frac{\mathrm m}{\mathrm s}}$

Plugging in both possible values of we found earlier, we get $t=0.862\,\mathrm s$ (if is positive) or $t=7.42\,\mathrm s$ (if is negative).

Answer from: Quest

i am pretty sure it is c because when people are closer together, they cannot move as freely. picture being in a mostly full elevator, for instance. the people are right next to you, and you can barely move.

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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The driver of a pickup truck going 89.9 km/h applies the brakes giving the truck a uniform decelerat...