X vs t graph for a car that's moving away from the origin and slowing down

[LenaS]

You're traveling at 25m/s when you notice something ahead. you apply the brakes and come to a stop in 4 seconds. construct a x vs.t graph, a v vs. t graph, and an a vs. t graph for the 4 sec time interval
if i could get what the position vs time graph looks like, i could get the rest. At first i had something that looked like an inverse graph for my position vs time, but i realized that would mean that the object was slowing down but approaching the origin. I have no clue how to show it slowing down but moving away from the origin.

 

[Dadface]

Hello lena,in your question you have two a vs t graphs,should one of these be an s(distance) vs t graph.I think it will be easier for you to start by sketching the v vs t graph.

 

[LenaS]

no, one is a velocity vs. time graph, one is a position vs. time graph, and the other is an acceleration vs. time graph.

I think i know what the v vs. t graph looks like. would it be a decreasing diagonal line from 25 m/s on the y-axis to 4.0 sec on the x?

 

[Dadface]

The acceleration is the slope of the velocity time graph(change of velocity/time)so the a vs t graph will be easy to sketch.When you say "position" then position with respect to what?I think by position the question is asking you to sketch a graph of distance from point where brakes were first applied against time.From a velocity time graph you can get the distance traveled from the area under the graph.

 

[LenaS]

wouldn't the area of the v vs. t graph give you the change in position though? I believe the position vs. time graph is asking the car's place at that specific time after the brakes were applied. I am not sure how to make the graph less steep(to show it's slowing down) yet go up away from the origin. Unless it's going towards the origin..?

 

[LenaS]

wait a minute! would a side opening parabola that flattens out in a straight line work?

 

[Dadface]

wouldn't the area of the v vs. t graph give you the change in position though? I believe the position vs. time graph is asking the car's place at that specific time after the brakes were applied.
How do you measure where a place is?The question wants you to find the distance from the point where the brakes were first applied.
I am not sure how to make the graph less steep(to show it's slowing down) yet go up away from the origin. Unless it's going towards the origin..?

.
You correctly described the velocity time graph.The velocity reduces from its initial value to zero in 4 seconds.It is a straight line of negative slope.Why not try the acceleration time graph next?

 

[LenaS]

the acceleration vs. time graph would just be a horizontal line on negative 5

 

[Dadface]

Horizontal yes,negative value yes but not negative 5.Change of velocity =25, time =4 and acceleration = change of velocity/time

 

[LenaS]

oh! right, sorry for some reason i was thinking the change of velocity was 20! so it'd be -6.25

 

[Dadface]

Correct.Now try a distance time graph.The distance traveled equals the area under the velocity time graph.To help you see this may I suggest that you find the total distance traveled in 1s,2s,3s and 4s?

 

[LenaS]

1 s - (25 x 1)/2 = 12.5 m
2 s - (25 x 2)/2 = 25 m
3 s - (25 x 3)/2 = 37.5 m
4 s - (25 x 4)/2 = 50 m

 

[LenaS]

so would the car's displacement be 50 m?

 

[Dadface]

I agree with the total distance traveled in 4s,this is the area under the whole graph which is the area of the whole triangle.Look again at your other values.In the first second,for example,the velocity drops not from 25 to zero but from 25 to 18.75.

And yes,the maximum displacement is 50 m but the graph is not a straight line

 

[LenaS]

1s - (6.25 x 1)/2 = 3.125 m
2s - (12.5 x 2)/2 = 12.5 m
3s- (18.75 x 3)/2 = 28.125 m

 

[Dadface]

Way off lena.The velocity reduces by 6.25 m/s every second so after 1s v=18.75,after 2s the velocity=12.5 after 3s v=6.25 and after 4s v=0.
Distance traveled in first sec=0.5(25+18.75)*1 etc
It's 11.40 in the UK so goodight.

 

[LenaS]

alright, thanks for your help!

 

[Dadface]

Good morning.Hello Lena did you manage to finish the problem?

 

[LenaS]

Good Afternoon(?)!
yes, i did - and i just got 100% on my quiz! :) Thanks so much!