YumYum247
  • YumYum247
how do i plot a velocity time graph from an average velocity of an object?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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YumYum247
  • YumYum247
if a runner accelerated along the sidewalk at 0.5m/sec2 for 20 sec. Assume that the runner started from rest position.....
anonymous
  • anonymous
You will plot velocity on the vertical (y) axis and the time on the horizontal (x) axis. All you have to do is plot the values and connect all of them by a line. Make sure velocity is given in meters per second (m/s) and time in seconds (s) unless instructed otherwise.
YumYum247
  • YumYum247
can you please show me how do i make the data table so i can plot the points neatly...

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YumYum247
  • YumYum247
|dw:1438454610769:dw|
YumYum247
  • YumYum247
how do i use the average velocity given in the question to plot the points on the graph.
anonymous
  • anonymous
Velocity (in m/s) is [distance in metres] divided by [time in seconds.]
YumYum247
  • YumYum247
so v = d/t
YumYum247
  • YumYum247
but how's that gunna help me?
YumYum247
  • YumYum247
if i want to know how far did the runner get in 7 sec, how do i figure that out???
YumYum247
  • YumYum247
@oldrin.bataku
anonymous
  • anonymous
if the acceleration is constant \(a\) then the total change in velocity from the start until time \(t\) is just \(v-v_0=at\); in our case, we starta t rest so \(v_0=0\)
YumYum247
  • YumYum247
yes the accelaeration is constant.
anonymous
  • anonymous
so the velocity is given by \(v=at\), so in terms of a velocity vs time plot we see that the graph is a simple line with slope \(a\) passing through the origin
YumYum247
  • YumYum247
i just need to plot how much distance he covered from rest to 20sec.
anonymous
  • anonymous
well, if we have a velocity vs time plot, then the distance covered between \(t=t_1\) and \(t=t_2\) is just the total area bounded by our velocity graph between those points:
anonymous
  • anonymous
|dw:1438456604986:dw|
YumYum247
  • YumYum247
can you please just do one for me, i'll do the rest myself.... can you figure out how much distance the runner covered in let's say 7 seconds????
anonymous
  • anonymous
|dw:1438456642313:dw|
anonymous
  • anonymous
so the distance they go between \(t_1,t_2\) is just the area under the graph between \(t_1,t_2\).
YumYum247
  • YumYum247
the area under the triangle is the displacement of the object....i get that...but
YumYum247
  • YumYum247
did you use Tf - Ti = a X t ?
YumYum247
  • YumYum247
because i need to make a data table to record all the points until 20 second and plot them on the velocity time graph.
anonymous
  • anonymous
okay, so here we have \(t_1=0\), \(t_2=7\), and our equation is again \(v(t)=0.5t\). so our total displacement is: $$A=\frac12 bh$$here our base is \(t_2=7\) since \(t_1=0\) and our height is \(v(t_2)=v(7)=3.5\) so: $$A=\frac12(7)(3.5)=12.25$$
anonymous
  • anonymous
you could also have done that work using an integral instead of manually finding teh area of a triangle
YumYum247
  • YumYum247
it doesn't look right because i have the answer key and the maximum displacement the runner makes under 20 sec is 10 meters.... i'm willing to show you the answer tho but i don't know how they got those points.....in the first place.
anonymous
  • anonymous
i dont know what points youre talking about
YumYum247
  • YumYum247
|dw:1438457170341:dw|
YumYum247
  • YumYum247
this is how it looks......
anonymous
  • anonymous
the kinematic equation gives \(x=x_0+v_0 t+\frac12 at^2\), here \(a=0.5,t=7\) and \(x_0=v_0=0\) so \(x=\frac12\cdot0.5\cdot7^2=12.25\)
anonymous
  • anonymous
actually, no it doesn't; you've misread it. that's the *velocity* against time, so it's going 10 meters *per second* in velocity after 20 seconds of time
YumYum247
  • YumYum247
i'm sorry it's a velocity time graph, the velocity/speed of the runner goes up to 10m/s under 20sec
anonymous
  • anonymous
yes, and that doesn't contradict the fact the runner goes 12.25 meters in 7 seconds.
YumYum247
  • YumYum247
ok
anonymous
  • anonymous
his speed after 7 seconds is only 3.5 m/s as I stated before
anonymous
  • anonymous
with an average speed of \((3.5-0)/2=1.75\) m/s, which is indeed correct: $$\frac{12.25}7=1.75$$ m/s
YumYum247
  • YumYum247
ok what formula did you use plase?????? :(
YumYum247
  • YumYum247
ok so i need to make a data table of this situation as drawn above, so what equation do i use to plot the speed of the runner every second????
anonymous
  • anonymous
since the velocity vs time graph is a line, the average velocity between \(t_1=0,t_2=7\) s is just the midpoint, hence \((3.5-0)/2=1.75\) m/s. this matches the alternative calculation of the average velocity as the total displacement \(12.25\) m divided by the time elapsed \(t_2-t_1=7 \) s, since \(12.25/7=1.75\) m/s as well
anonymous
  • anonymous
the equation for the velocity at time \(t\) is just \(v=0.5t\), since the acceleration is constant. this is the straight line we've been graphing
YumYum247
  • YumYum247
ok let me try on my own and i'll ask you to check my work. Thank you!!! :")
YumYum247
  • YumYum247
omg i see how you got 10m....LOL
YumYum247
  • YumYum247
ok one more thing, if the average velocity of an object is 0.5m/sec2 how quick or how fast will it accelerate every second.

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