Question

how do you get distances form a velocity time graph???

Answer 1

Integrate.

Answer 2

from the graph drop perpendiculars onto the time axis

Answer 3

this will give u a geometric figure (in case of uniform acceleration) like triangle, trapezium..
depending on the figure, use the formula for calculating the area of that figure....
the figure got for area gives the distance travelled..

Answer 4

so how would you do it for qs 3??
http://monmouth.monmoodle.co.uk/file.php?file=%2F187%2FMEI%20A-LEVEL%20HOMEWORK%20RESOURCES%2FM1%2FM1%20Multiple%20Choice%2FMC%20Motion%204.pdf

Answer 5

cause i cant really see a traingle or trapezium :S

Answer 6

u mean the question about the average speed???

Answer 7

yes qs 3 and 2 please :)
thanks

Answer 8

actually dont worry bout qs 3

Answer 9

if u drop a perpendicular from the first bend (at t=3s) u get a triangle with base = 3 and height = 2
so this are = 1/2*base*height = 3

Answer 10

then drop another perpendicular onto the time axis where the flat line ends (at t=6s)
now with the previous perpendicular at t=3 and now with perpendicular at t=6, you get a rectangle with length = 3 (6-3) and width = 2
its area = l * w = 3 * 2 = 6

Answer 11

finally drop a perpendicular from the end of the climbing line (at t=10s) and you get a trapezium with parallel sides of size 2 and 10 and its height = 4 (10-6)

area of trapz = h/2* (sum of parallel sides)
= 4/2 * (2 + 10)
= 2*12 = 24

Answer 12

so adding all three 3+ 6 + 24 = 33

so the answer to question 2 is 33 m

Answer 13

did u get this???

Answer 14

i got 32
lemme see where i went wrong :S
thanks for the detailed explanation :)

Answer 15

you know for the trapezium???
where did you get 4/2 form??

Answer 16

see for the trapezium, its height will be the difference on the time axis 10-6 = 4
so h/2 = 4/2

ill just send u a file

Answer 17

THANK YOU I GOT IT :)

Answer 18

here it is ....

Answer 19

just open and see the markings......