Question

an object moves along the x axis according to the equation x = 3.00t^2 - 2.00t + 3.00 where x is in meters and t is in seconds. Determine the average speed between t = 2.00 s and t = 3.00s

do i just substitute t again? @mathslover :) just help if you can hehe

Answer 1

@lgbasallote The average speed is the distance passed divided by the time it took :

At T=2 the object was at 3*2^2-2*2+3 = 11m .
At T=3 the object was 3*3^2-2*3+3 = 24m.

Total distance passed is 24-11 = 13 meters. It took one second . So the average speed is 13/1 = 13m/s.

Answer 2

that is the answer...

Answer 3

distance passed is just substituting t right?

Answer 4

what about for instantaneous acceleration??

Answer 5

The acceleration function is the derivative of the speed function(which we found in B).
So :
a(t) = v'(t) = 6.
this function is constant , so the instantaneous acceleration is always 6m/s^2 at any time.

Answer 6

uhh is there a non-calculus way??

Answer 7

the only way to find the acceleration is differentiating the velocity.

Answer 8

wait..i skipped a letter...what is the instantaneous speed at t = 2 and t = 3?

is it \[\frac{3(2)^2 - 2(2) + 3}{2}\] for t=2?

Answer 9

and acceleration = \(\frac{d^2r}{dt^2} (3t^2 - 2t + 3)\) right?

Answer 10

i mean you got the acceleration by taking the double derivative of 3t^2 - 2t + 3?

Answer 11

Yes, the when differentiating the displacement you get the velocity then you differentiate the velocity to get the acceleration.

Answer 12

@mathslover did i do my instantaneous speed right?

Answer 13

Instantaneous speed at time t is the derivative of the location function at the point t.

so we need the derive the function : x(t) = 3*t^2-2*t+3
so v(t) = x'(t) = 6t-2 .
for the time t=2 we get 6*2-2 = 10m/s .
for the time t=3 we get 6*3-2 = 16m/s

Note that the answer for A is actually the average of the two answers of B . That is because the acceleration here is constant but that is not always the case! That is why the form to solve A is more general and will always give you the correct result even when this is not the case.

Answer 14

do i ALWAYS use derivatives??

Answer 15

physics can include derivative in any question

Answer 16

that is why CALCULUS is the must part of maths and physics
if no maths than no physics :(

Answer 17

hmm instantaneous acceleration for t = 2 and t = 3 are both 6 right? since a is constant

Answer 18

very great .... congr8s u r going right

Answer 19

last follow up question is...what time is the object at rest?

i assume i equate 3t^2 - 2t + 3 = 0

then solve for t?

Answer 20

oh wait!!!

6t - 2 = 0!!

Answer 21

because at rest means velocity is 0

Answer 22

right...

Answer 23

so when t = 1/3 seconds the object is at rest?

Answer 24

right !!

Answer 25

yay thanks!

Answer 26

:) my pleasure.... yw:) ....mention not ....and thanks for tagging me here :)