Question

: If a body travels 0.2m in the first 0.2s and 0.22m in the next 0.4s , what will be its velocity at the end of the 0.7th second ?

Answer 1

are we to assume a constant acceleration?

Answer 2

given to points to work from, the simplest thing you can get is a linear function

Answer 3

( t, m)
( .2, .2)
-(.4,.22)
-------
-.2, -.02; slope between them is 1/10

Answer 4

y = .1t -.1(.2) +.2

y = .1t -.02 +.2

y = .1t + .18

when t = .7 we get:

y = .1(.7) + .18 = .25

maybe

Answer 5

these are positions tho, and not velocity.

the velocity itself is just .1m/s i believe

Answer 6

but im reading it wrong again i believe .....

If a body travels 0.2m in the first 0.2s and

so at .2 we have a position of .2m

0.22m in the next 0.4s

at .6s we have a position of .22m or an extra .22m?

Answer 7

still, given 2 data points, the best we can assume is a linear model and the velocity at any given point is the same as the slope of the line

Answer 8

but then again we have an extra point dont we, i assume at time 0s we are at a velocity of 0

Answer 9

if we are measureing it from rest then we need to fit a curve to 3 points: (0,0), (.2,.2), and (.6,.42) perhaps

Answer 10

our curve fits are then, if im reading the information correctly:

a0^2 + b0 + c = 0
a.2^2 + b.2 + c = .2
a..6^2 + b.6 + c = .42

c = 0


.04a + .2b = .2
.36a + .6b = .42

this turns into a doable system of equations by applying cramers rule:

b = .42(.04)-.36(.2) / .6(.04)-.36(.2)
= 1.15

a = .42(.2)-.6(.2) / .36(.2)- .6(.04)
= -.75

so our curve should be:

-.75 t^2 + 1.15 t ; let me dbl chk this with the wolf tho ... the wolf like it

Answer 11

with this as the position equation, we can take the derivative to determine the instanteous velocity at any given point.

v(t) = -1.5 t + 1.15

v(.7) = .1 m/s

Answer 12

i think you can work this out in terms of the average velocity of the body in each interval (assumong constant acceleration) we know s=d/t so therefore average speed for first 0.2 seconds= 1m/s therefore its maximum speed must be 2m/s (at 0.2s, at 0s v=o). now fro the second interval, average speed = 2.2/0.2=11 m/s since you started at 1m/s and get an average of 11 velocity at 0.4s =21m/s. therefore the acceleration is (21-2)/0.2=95m/s^2. therefore at 0.7s v= 95x0.7=66.5m/s. this seems right to me but the previous answer is just as likely to be correct if not more so considering i'm the one giving my solution :p.