Question

if a ball is thrown into the air with a velocity of 40ft/s, its height in feet t seconds later is given by y=40t-16t^2. find the average velocity for the time period beginning when t=2 and lasting 0.5 seconds

Answer 1

y = the upward distance = 40t - 16t^2 distance/time = velocity
Let V be the velocity so V = y /t = 40 - 16 t, t = 2
V = 40 - 32 = 8

Answer 2

ok wait

Answer 3

this is the equation
y=40t-16t^2

Answer 4

how did u plug t=2 and (0.5)seconds

Answer 5

yes, y = 40t - 16t^2
y /t is velocity = 40 - 16t st t = 2, v = 8

Answer 6

ok u factor

Answer 7

Vavg = (yf-yi)/(tf-ti)
yf = 40(2.5)-16(2.5)^2 = 0
yi = 40(2)-16(2)^2 = 16
tf = 2.5
ti = 2
Vavg = (0-16)/(2.5-2) = -16/.5 = -32 ft/s

Answer 8

what is tha 2.5

Answer 9

The problem asked for AVERAGE velocity from t=2 and lasting .5 seconds, so the initial time is t=2 and the final time is t=2.5

Answer 10

ok
how did u get 2.5 u added then

Answer 11

yes, if you start at t=2 and end .5 seconds later, then the final time is t=2.5

Answer 12

cool

Answer 13

p.kanan found the velocity when t=2, which is not the average velocity from t=2 to t=2.5 as the problem asked.

Answer 14

ohh ok

Answer 15

thank u quantish

Answer 16

do you understand? there is a difference between average velocity and velocity at a certain instant.

Answer 17

sip
another qs. in the denominator
is no supose to be
40(2)-16(0.5)^2

Answer 18

I really don't understand what you're asking.
what denominator?
your change in time i.e. time final - time initial belongs in the denominator if you're are finding Vavg.

Answer 19

Change in position i.e. y final - y initial belongs in the numerator.

Answer 20

ok