Physics, 04.09.2020 02:01, leandrogarin37p2g5ds

An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph for the values of t that are separated by the step Δt = 2s.

An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw t

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answered: dkdk31

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

$v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

$v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt$

Where:

$v_{a}$ , $v_{b}$ - Initial and final velocities, measured in meters per second.

$t_{a}$ , $t_{b}$ - Initial and final times, measured in seconds.

a(t) - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

$v_{4} = 2\,\frac{m}{s} +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)$

$v_{4} = -6\,\frac{m}{s}$

Region II (t = 4 s to t = 6 s)

$v_{6} = -6\,\frac{m}{s} +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt$

$v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)$

$v_{6} = -4\,\frac{m}{s}$

Region III (t = 6 s to t = 10 s)

$v_{10} = -4\,\frac{m}{s} +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt$

$v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)$

$v_{10} = 4\,\frac{m}{s}$

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw th

answered: matthewsorrow02

The velocity of a body under constant acceleration increases steadily with time

Please find attached the required velocity graph for values of t that are separated by Δt = 2s

The reasons the attached graph is correct are given as follows:

At t = 0 seconds

The initial velocity of the object at t = 0 is v = 2.0 m/s

The first point on the graph is (0, 2.0)

From t = 0 s, to t = 4 s

The acceleration from t = 0, to t = 4, a₁ = -2 m/s²

The velocity at t = 4 s, v₂ = 2.0 + (-2)×4 = -6

Therefore, the next point on the graph is (4, -6)

From t = 4 s to t = 6 s

From t = 4 to t = 6, the acceleration, a₂ = 1 m/s²

Therefore, v₃ = -6 + 1 × 2 = -4

The third point on the velocity graph is (6, -4)

From t = 6 s to t = 10 s

From t = 6 s to t = 10 s, we have, the acceleration, a₃ = 2 m/s²

The velocity, v₄ = -4 + 4 × 2 = 4

Therefore, the fourth point on the velocity graph is (10, 4)

With the above points, the velocity graph can be plotted using MS Excel

Learn more about plotting velocity time graph, here:

link

answered: Guest

howdy howdy!

i'd say it would be galileo!

answered: Guest

$\frac{2y^2-5y-7}{6y^2+10y+4}-\frac{7}{9y^2+6y}=\frac{6y^2-21y-14}{6y\left(3y+2\right)}$

step-by-step explanation:

we are given

$\frac{2y^2-5y-7}{6y^2+10y+4}-\frac{7}{9y^2+6y}$

we have to simplify it

firstly, we will factor it

$2y^2-5y-7=\left(y+1\right)\left(2y-7\right)$

$6y^2+10y+4=2\left(y+1\right)\left(3y+2\right)$

$9y^2+6y=3y(3y+2)$

$=\frac{\left(y+1\right)\left(2y-7\right)}{2\left(y+1\right)\left(3y+2\right)}-\frac{7}{3y(3y+2)}$

now, we can cancel like terms

$=\frac{\left(2y-7\right)}{2\left(3y+2\right)}-\frac{7}{3y(3y+2)}$

now, we can combine it

$=\frac{3y(2y-7\right)}{6y\left(3y+2\right)}-\frac{7\times 2}{6y(3y+2)}$

$=\frac{3y(2y-7\right)-14}{6y\left(3y+2\right)}$

$=\frac{6y^2-21y-14}{6y\left(3y+2\right)}$

so, we get

$=\frac{6y^2-21y-14}{6y\left(3y+2\right)}$

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An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw th...