Question

I have the answer can someone check me please...
Hugh is driving a car down the road. Let d(t) give the distance he is from home t hours after he left.
If d(t)=-1.5t^2+64t find Hugh's acceleration 4 hours after he left
Acceleration is?

Answer 1

Acceleration is the second derivative of displacement.
What did you get for an answer farmgirl? c:

Answer 2

I got 280

Answer 3

So it seems you found Hugh's `distance` after 4 hours.
\(\large d(4)=-1.5(4)^2+64(4)=232\)
There was a small mistake in your calculation I think. Doesn't matter though, we don't want the distance, we want the acceleration at 4 hours.

\[\large d'(t)=v(t) \qquad \rightarrow \qquad v(t)=-3t+64\]The second derivative will be \(\large d''(t)=a(t)\), the acceleration function that we're looking for. Think you can find the second derivative? After you do, plug in \(\large t=4\).

Answer 4

so then is the answer 52

Answer 5

No, it appears you plugged 4 into the velocity function.
Take the derivative again to get the acceleration function, and then plug 4 into it.

Answer 6

this is a calculus problem. Do you understand how to take the second derivative ?

Answer 7

I am confussed

Answer 8

can you take the derivative of
d(t)= -1.5t^2+64t
with respect to t ?

zepdrix already did it for you , and got
v(t) = -3t + 64

now you have to take the derivative of v(t) with respect to t
can you do that ?

Answer 9

-3t+64t

Answer 10

ok that is the first derivative. now take the derivative of that

Answer 11

how do you do that

Answer 12

first it is -3t+64 (not -3t+64t)

the derivative of -3t is ?
and derivative of 64 is ?

Answer 13

\[ \frac{d}{dt} (-3t) = ?\]

Answer 14

-12 and 256

Answer 15

do you know how to take the derivative of -3t where t is the variable ?

Answer 16

do you know how to do
\[ \frac{d}{dx} 3x \]

Answer 17

If not, you really can't do this problem.

Answer 18

wait just redid it -85.33

Answer 19

then 1024

Answer 20

Is that right

Answer 21

I guess you don't understand. To do this problem you take the derivative of
d(t)= -1.5t^2+64t
you get
v(t) = -3t + 64 (this says d distance/dt = velocity)

the change in velocity with respect to time is acceleration
a(t)= dv/dt = -3

that says the acceleration is always -3 (what ever units they are using)

Answer 22

If that does not make sense, you have to brush up on your calculus
(and learn that velocity is the first derivative of distance, and acceleration is the 2nd derivative of distance)

Answer 23

See http://www.khanacademy.org/math/calculus/differential-calculus/derivative_intro/v/calculus--derivatives-1--new-hd-version

Answer 24

ok thanks