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YumYum247
 one year ago
how do i plot a velocity time graph from an average velocity of an object?
YumYum247
 one year ago
how do i plot a velocity time graph from an average velocity of an object?

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YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1if a runner accelerated along the sidewalk at 0.5m/sec2 for 20 sec. Assume that the runner started from rest position.....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You 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
 one year ago
Best ResponseYou've already chosen the best response.1can you please show me how do i make the data table so i can plot the points neatly...

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1dw:1438454610769:dw

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1how do i use the average velocity given in the question to plot the points on the graph.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Velocity (in m/s) is [distance in metres] divided by [time in seconds.]

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1but how's that gunna help me?

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1if i want to know how far did the runner get in 7 sec, how do i figure that out???

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if the acceleration is constant \(a\) then the total change in velocity from the start until time \(t\) is just \(vv_0=at\); in our case, we starta t rest so \(v_0=0\)

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1yes the accelaeration is constant.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 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
 one year ago
Best ResponseYou've already chosen the best response.1i just need to plot how much distance he covered from rest to 20sec.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well, 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1438456604986:dw

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1can 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
 one year ago
Best ResponseYou've already chosen the best response.0dw:1438456642313:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the distance they go between \(t_1,t_2\) is just the area under the graph between \(t_1,t_2\).

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1the area under the triangle is the displacement of the object....i get that...but

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1did you use Tf  Ti = a X t ?

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1because i need to make a data table to record all the points until 20 second and plot them on the velocity time graph.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, 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
 one year ago
Best ResponseYou've already chosen the best response.0you could also have done that work using an integral instead of manually finding teh area of a triangle

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1it 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
 one year ago
Best ResponseYou've already chosen the best response.0i dont know what points youre talking about

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1dw:1438457170341:dw

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1this is how it looks......

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 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
 one year ago
Best ResponseYou've already chosen the best response.0actually, 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
 one year ago
Best ResponseYou've already chosen the best response.1i'm sorry it's a velocity time graph, the velocity/speed of the runner goes up to 10m/s under 20sec

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, and that doesn't contradict the fact the runner goes 12.25 meters in 7 seconds.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0his speed after 7 seconds is only 3.5 m/s as I stated before

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0with an average speed of \((3.50)/2=1.75\) m/s, which is indeed correct: $$\frac{12.25}7=1.75$$ m/s

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1ok what formula did you use plase?????? :(

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1ok 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
 one year ago
Best ResponseYou've already chosen the best response.0since the velocity vs time graph is a line, the average velocity between \(t_1=0,t_2=7\) s is just the midpoint, hence \((3.50)/2=1.75\) m/s. this matches the alternative calculation of the average velocity as the total displacement \(12.25\) m divided by the time elapsed \(t_2t_1=7 \) s, since \(12.25/7=1.75\) m/s as well

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the 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
 one year ago
Best ResponseYou've already chosen the best response.1ok let me try on my own and i'll ask you to check my work. Thank you!!! :")

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1omg i see how you got 10m....LOL

YumYum247
 one year ago
Best ResponseYou've already chosen the best response.1ok 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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