Question

a. What is the speed at t=20.0sec?
b. Find the distance traveled in the first 20 seconds

Answer 1

here's the graph

Answer 2

what is teh acceleration at t=20?

Answer 3

lol, read it a little off ...

start at the begining; let v = at

when a=0, then v remains constant

Answer 4

something akin to that ... but my last leg might be off

Answer 5

v = 2t, from 0 to 10

v = 2(10), from 10 to 15

v = -3t+c, from 15 to 20

v = -3(20)+c from 20 onwards

Answer 6

What does C represent?
And for the second part of my question, I got the answer 237.5, and the website I'm using to submit my assignment says it is within 10% of the correct answer, so I'm having trouble fixing my mistake. I found the area underneath each part of hte graph, 100 from 0-5 sec, 100 from 10 to 15 sec, and 37.5 from 15-20sec

Answer 7

some concepts first:

we agree that v = at ?

Answer 8

c just represents the shift by starting at 15 instead of 0

Answer 9

we could just as well consider this in three parts;

v = 2t, from 0 to 10

v = 2(10), from 0 to 5

v = 20 - 3t, from 0 to 5

Answer 10

each part is calibrated to start the count anew is all

Answer 11

the displacement is just the areas of the 3 different parts of velocity

triangle: 10x20
rectangle: 20x5
rectangle-ish: (20+5)/2 x 5

Answer 12

182.5 displacement seems good to me

Answer 13

forgot how to do math .... lol

Answer 14

100+100+5(25)/2 = 262.5

Answer 15

ok that's what I got lol!

Answer 16

Can we go back to the speed at t=20?

Answer 17

t=20 is the end of the third leg of this which i simplified as:

v = 20 - 3t, from 0 to 5 since it only lasts 5 seconds and it start at 20

Answer 18

So i it would be 5? I was thinking you would do v+at with a=-3 and t=5

Answer 19

the trip starts at v=0 a=2, from 0 to 10 seconds; v = 0 to 20

for the next 5 seconds they are just cruising along at 20 ..

now they start to de-accelerate by -3: 20 - 3t for the next 5 seconds; ending in: 20-3(5) = 5 for a final velocity

Answer 20

poor choice of words ... de-accelerate at 3 or accelerate at -3 ...

Answer 21

Oh! I understand now thank you!

Answer 22

youre welcome