Physics, 21.12.2020 09:10 saifulcrc468

A bus travels at a constant speed. it stops for a short time and then travels at a higher constant speed. draw a distance-time graph for this journey.

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Answer from: yurimontoya759

See the explanation below

Explanation:

Since the bus travels at a constant speed, we can use the following equation.

$x=x_{o}+v*t$

where:

x = final position [m]

xo = initial position = 0 (if the bus starts from some reference point)

v = constant velocity [m/s]

t = time [s]

Let's take some data as an example and then graph them.

v = 2 [m/s]

t = 2 [s]

That is, after two seconds, the final position will be:

$x = 0 + 2*2\\x = 4 [m]$

After the two seconds, the bus stops for about 2 seconds more. Then the bus will go with a velocity of 4 [m/s] for two seconds more.

So the total time is 6 [s].

Let's draw the x-t graph.

We see that the position changes during the first two seconds, from a position 0 to 4 [m]Then the bus stops for two seconds, that is, for 4 seconds the bus has traveled 4 meters.Then the bus moves to 4 [m/s], for 2 more seconds. That is, the bus will have moved up to 6 seconds.

And the final offset must be calculated using the same equation.

$x = x_{0}+v*t\\x = 4+4*(6-4)\\x = 4 +8\\x = 12 [m]$

A bus travels at a constant speed. it stops for a short time and then travels at a higher constant s

Answer from: Quest

Dis the

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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