Question

Physics,
Can you help me with (C) please.
https://www.dropbox.com/s/f9v9w4z5fk9u2vz/Screenshot%202015-11-15%2000.44.17.png?dl=0

Answer 1

Do you have a little understanding for integration from calculus or is this Physics class algebra based?

Answer 2

Integrate velocity to find displacement.

Answer 3

Which basically means find the area under the curve of each portion of the graph to find the total area which is the total displacement.

Answer 4

how about I use V = V_0 + at

Answer 5

remember I only want part (C)

Answer 6

the one asking for accel

Answer 7

so there is this intepretation https://www.dropbox.com/s/j358aun6hma5fkl/Screenshot%202015-11-15%2003.13.30.png?dl=0

Answer 8

and then there is mine: working with the formula V = V_0 + at

Answer 9

However both give different result. So. who's correct and why ?

Answer 10

@zepdrix

Answer 11

you should find the equation for the speed from 4 seconds to 6 seconds, and then take the derivative of that function. And evaluate it at t=5sec

Answer 12

That is to find the instantaneous acceleration.

Answer 13

how about V = V_0 + at

Answer 14

That equation is to find final velocity as acceleration causes the object to speed up or slow down as time progresses with an initial velocity.

Answer 15

Which parts do you need hep with at this point?

Answer 16

I would assume you have for the least done a and b, is that so?

Answer 17

@SolomonZelman, Please let me know if my replies were incorrect so I can learn from this also.

Answer 18

(c)
Acceleration is the rate at which velocity is gained or lost, or the second derivative of position, or first derivative of velocity ──> (the sope of velocity)... can be defined in various ways.

So, acceleration in this case would simply be the instantaneous slope
(just line, so you don't need calculus).

(d)
Average velocity would be the average value of the function.

(e)
slope of secant between t=1 and t=5.

Answer 19

Laggy today :(

Answer 20

i though I could use V = V_0 + at and solve for a to find a (PART C)

Answer 21

Well, you are given the function, and the acceleration at x=c is the instantaneous slope at x=c. In a case of a line you can easily find this slope

Answer 22

I mean instantaeous slope of velocity.

Answer 23

\(\large\color{black}{ \displaystyle A(x)= V'(x) }\)
\(\large\color{black}{ \displaystyle A(5)= V'(5) }\)

A - acceeration
V - velocity

Aternatively,
\(\large\color{black}{ \displaystyle A(5)= }\) the slope at t=5

Answer 24

I don't think it would be valid in this case to use the formula you proposed, unless you redefine things.

Define t=4 as t=0
and equivalently, t=5 as t=1

Then,

V = V_0 + at
(where V denotes the final velocity)

V_0 = V(4)=14
V = V(5)=7

14 = 7 + a•(5-4)

Answer 25

You can't use it just

V = V_0 + at
V(5) = V(0) + a•5
7 = 0 + a•5
7/5 = a

but that would be INCORRECT.

Answer 26

what I did was basicaly defined the function as: