Question

Find the distance, in meters, a particle travels in its first 10 seconds of travel, if it moves according to the velocity equation v(t)= 49 − 9.8t (in meters/sec).

A: −49
B: 0
C: 245
D: 441

Answer 1

@primeralph can u plz help? :)

Answer 2

@campbell_st

Answer 3

i don't really see any question

Answer 4

you just stated the info

Answer 5

o o o i take it back lol
you did have the question there

Answer 6

its ok :)

Answer 7

okay let's see what do we have here

Answer 8

ok so \[\Delta x=49t-4.9t^2 \]
evaluate at t=10

Answer 9

0

Answer 10

triangle x = 0

Answer 11

this is one of the position equations
you should have something similar to this in your notes

Answer 12

what are you talking about?

Answer 13

i plugged in t= 10 into the deltax = 49t-4.9t^2

Answer 14

this is what i meant \[\Delta x =4.9(10)-4.9(10)^2\]

Answer 15

it's not zero

Answer 16

49-490=-441

Answer 17

So it would be: D, 441! thank u!

Answer 18

hmm the answer you are given is positive
the one we found is negative something is off

Answer 19

was you equation
\[v(t)=4.9-9.8t\]

Answer 20

your?

Answer 21

v(t)= 49 − 9.8t (in meters/sec).

Answer 22

i can't see what you wrote lol

Answer 23

its v(t)=49-9.8t

Answer 24

no decimal between 4 & 9 :)

Answer 25

you sure?

Answer 26

100%

Answer 27

okay then you have \[\Delta x=0\]

Answer 28

i see why you said that from the start haha

Answer 29

perfect :) thank u

Answer 30

i have 1 more question do u mind helping me with it?

Answer 31

np!

Answer 32

let's see

Answer 33

For an object whose velocity in ft/sec is given by v(t) = −3t^2 + 5, what is its displacement, in feet, on the interval t = 0 to t = 2 secs?

A: 6.607
B: 2
C: −2.303
D: 2.303

Answer 34

I know its not A, 6.607 I think its C, -2.303

Answer 35

what did you do

Answer 36

http://openstudy.com/users/mathrulezz#/updates/54ed39c9e4b0dc552b644449

Answer 37

did you compute the integral

Answer 38

that should gave the desplacement

Answer 39

Just integrate.

Answer 40

can u please walk me through the steps

Answer 41

do you know about integrals
you are doing physics I yes?

Answer 42

calc 1 :)

Answer 43

well you should know how to integrate unless you just started calc1

Answer 44

at any rate let's do it
\[\int_{0}^{2}(-3t^2+5 )dx=\left[-t^3+5t\right]_{0}^{2}=-(2)^3+5(2)+0-0\]

Answer 45

-8+10+0-0=2 :)

Answer 46

brilliant thank u

Answer 47

that's the use of th fundamental theorem of calc

Answer 48

:) np!

Answer 49

i got this 1 wrong :(
Find the distance, in meters, a particle travels in its first 10 seconds of travel, if it moves according to the velocity equation v(t)= 49 − 9.8t (in meters/sec).

Answer 50

i don't think that's 49 should be 4.9 i guess

Answer 51

you can use the integral once more with that as well

Answer 52

got it :) thank u

Answer 53

For an object whose velocity in ft/sec is given by v(t) = −t2 + 6, what is its displacement, in feet, on the interval t = 0 to t = 3 secs?

9.000
−9.000
10.596
−3.00

Answer 54

they all the same problems my friend
just follow the same procedure to answer them

Answer 55

just integrate again from 0 to 3

Answer 56

can you please walk me through those steps again

Answer 57

i know its either A,9.000 or C,10.596

Answer 58

@mathrulezz
The first question about the particle requires you to find the distance travelled, not displacement.
So it is not
displacement=\(x_{final} - x_{initial}\)
It is total distance travelled.
See diagrams below:

So the particle has travelled to x=122.5 and back, for a total \(distance\) of 245.