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

1. YumYum247

if a runner accelerated along the sidewalk at 0.5m/sec2 for 20 sec. Assume that the runner started from rest position.....

2. 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.

3. YumYum247

can you please show me how do i make the data table so i can plot the points neatly...

4. YumYum247

|dw:1438454610769:dw|

5. YumYum247

how do i use the average velocity given in the question to plot the points on the graph.

6. anonymous

Velocity (in m/s) is [distance in metres] divided by [time in seconds.]

7. YumYum247

so v = d/t

8. YumYum247

but how's that gunna help me?

9. YumYum247

if i want to know how far did the runner get in 7 sec, how do i figure that out???

10. YumYum247

@oldrin.bataku

11. 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 $$v_0=0$$

12. YumYum247

yes the accelaeration is constant.

13. 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

14. YumYum247

i just need to plot how much distance he covered from rest to 20sec.

15. 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:

16. anonymous

|dw:1438456604986:dw|

17. 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????

18. anonymous

|dw:1438456642313:dw|

19. anonymous

so the distance they go between $$t_1,t_2$$ is just the area under the graph between $$t_1,t_2$$.

20. YumYum247

the area under the triangle is the displacement of the object....i get that...but

21. YumYum247

did you use Tf - Ti = a X t ?

22. 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.

23. 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: $$A=\frac12(7)(3.5)=12.25$$

24. anonymous

you could also have done that work using an integral instead of manually finding teh area of a triangle

25. 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.

26. anonymous

i dont know what points youre talking about

27. YumYum247

|dw:1438457170341:dw|

28. YumYum247

this is how it looks......

29. 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 $$x=\frac12\cdot0.5\cdot7^2=12.25$$

30. 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

31. YumYum247

i'm sorry it's a velocity time graph, the velocity/speed of the runner goes up to 10m/s under 20sec

32. anonymous

yes, and that doesn't contradict the fact the runner goes 12.25 meters in 7 seconds.

33. YumYum247

ok

34. anonymous

his speed after 7 seconds is only 3.5 m/s as I stated before

35. 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

36. YumYum247

ok what formula did you use plase?????? :(

37. 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????

38. 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

39. 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

40. YumYum247

ok let me try on my own and i'll ask you to check my work. Thank you!!! :")

41. YumYum247

omg i see how you got 10m....LOL

42. 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.