Recent content by LenaS

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

alright, thanks for your help!

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

so would the car's displacement be 50 m?

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

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

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

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

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...

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?

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...