Question

A particle moving in a straight line has a velocity given by v(t)=2sin(t)-1. What is the rate of change of velocity, called the average acceleration, of the particle over the interval (π/6)
≤ t ≤ (π/4)?

Answer 1

do you know how to find the derivative of that?

Answer 2

one second

Answer 3

that's what they are asking you to do.

Answer 4

the derivative of velocity makes your acceleration function

Answer 5

so is it 2cos(t)-1?

Answer 6

one second

Answer 7

i could be wrong, but i think it is just cos(t)

Answer 8

I believe the derivative would you the "instantaneous" acceleration.
what you are being asked to find is the "average" acceleration.

Answer 9

or 2cos(t)

Answer 10

average acceleration = (final_velocity - initial_velocity) / time_taken

Answer 11

so i dont need to find the derivative asnaseer? and ya i was wrong, the derivative would be 2cos(t)

Answer 12

would the derivative over the interval give you the same answer?

Answer 13

bleh123: thats right - no need to find derivatives

Answer 14

so what do i do with the interval it gives?

Answer 15

he doesn't have time though.

Answer 16

*she

Answer 17

time_taken = start_time - end_time

Answer 18

sorry - other way around

Answer 19

ah ok

Answer 20

start_time = \(\pi/6\)
end_time = \(\pi/4\)

Answer 21

umm so how would i solve this?

Answer 22

I have given you all the information you need to work it out

Answer 23

would initial velocity be 0?

Answer 24

yes - that looks right

Answer 25

the equation is pretty ugly.

Answer 26

2sin(t)-1 over pi/6 -pi/4

Answer 27

initial_velocity = \(v(\pi/6)\)
final_velocity = \(v(\pi/4)\)

Answer 28

-_- so i set it up wrong?