Question

Car A has a positive velocity v(t) as it travels along a straight road, where v is a differentiable function of t. The velocity of the car is recorded for several selected values of t over the interval 0<=t<=60 seconds, as shown in the table below

t (seconds): 0 10 20 30 40 50 60
v(t) (rate) : 5 14 7 11 12 40 44

Answer 1

Car B travels along the same road with an acceleration of a(t) = 1/(t+9)^1/2 ft/sec2
At time t=0 seconds, the velocity of Car B is 3ft/sec. Which car is traveling faster at t=40 seconds?

Background information that may or may not pertain to the question:
Car A travels 1300 feet
The acceleration at t=25 is 5/2

Answer 2

I found the acceleration of Car A from 0 to 40, which was 40/7

I also found the acceleration of Car B from 0 to 40, which was -14
I got it by plugging in a(40), which was 1/7, for the equation and using that as my f(40)

Am I on the right track? I got that Car A was faster but I don't think my process is right.

Answer 3

did u find the velocity of Car B ?

Answer 4

No I didn't. Would I need to integrate 1/(t+9)^1/2 to do that?

Answer 5

use below :-

velocity of Car B = \(\large \int a(t) dt\)

Answer 6

yess

Answer 7

and use the initial value (t=0, v=3) to find the constant

Answer 8

Do I find the integral from 0 to 40?

Answer 9

we can find the indefinite integral,
that gives u velocity function

Answer 10

I got 2(t+9)^1/2 + C, but this is a definite integral

If I find it from 0 to 40, I get

14-0 which is 14

Answer 11

So if v(40) = 14, then . . .

40 - 0 / 14 - 3

= 40/11 which is actually smaller than 40/7, so Car A is faster.

Does that look right to you?

Answer 12

wait a sec,

definite integral between 0 and 40 gives u area under acceleration curve,
wat does that represent ?

Answer 13

Ah the distance I believe. What am I trying to find then?

Answer 14

you wanted to find velocity at t=40

Answer 15

The velocity at t=40 was found from integrating the acceleration though right?
Maybe the integral is not from 0 to 40, but when I plug 40 in for v(40) I get 14 right?

Answer 16

At time t=0 seconds, the velocity of Car B is 3ft/sec. Which car is traveling faster at t=40 seconds?

\(\large v(t) = \int a(t) dt = \int 1/\sqrt{t+9} dt = 2 \sqrt{t+9} + c\)

apply initial values : \(\large v(0) = 3\)
\(\large 3 = 2\sqrt{0+9} + c \)
\(\large c = -3\)

velocity function :
\(\large v(t) = 2 \sqrt{t+9} -3\)

Answer 17

plugin t = 40, to find velocity at t=40

Answer 18

So v(40) = 11

Then 40 - 0 / 11 - 3 = 40/7, which is equal to car A.

Answer 19

nope, car A velocity at t=40 is 12

Answer 20

we simply pick the value from table directly

Answer 21

we are dealing wid 'instantaneous velocities'...
not average...

Answer 22

Okay let's see.
OH that makes sense
Okay well thank you very much. It was a little confusing at first but you've cleared it up. Thank you!

Answer 23

np :)