Question

Suppose that the equation of motion for a particle (where S is in meters and t in seconds) is
S=5t^3-7t
(a) Find velocity & acceleration as functions of t
velocity at time t......... 15t^2-7
acceleration at time t.. ?
(b) Find the acceleration after 1 second= ? (would I just plug in 1 from the equation in part a, after finding acceleration ?)
(c) Find the acceleration at the instant when the velocity is 0 ?

Answer 1

do you know calculus?

Answer 2

I'm barely learning

Answer 3

good enough :-)

velocity is the first derivative of position.
acceleration is the first derivative of velocity, or the second derivative of position

Answer 4

\[v(t) = \frac{d}{dx}[15t^2-7] =\]
\[a(t) = \frac{d}{dx}[v(t)] = \]

Answer 5

@whpalmer4 Oh okay so for acceleration it would be 15t
for part b I just plug in t=1 into 15t and get 15
for part c I would plug in 0 for the derivative I found for velocity ? and get -7 ?

Answer 6

oh, not so fast…

\[v(t) =\]?

Answer 7

v(t)=15t^2 -7

Answer 8

No, \(s(t) = 15t^2-7\)

Answer 9

\[v(t) = \frac{ds}{dt} = \]

Answer 10

oh, sorry, I misread your question…hold on

Answer 11

I'm confused my original equation was s=5t^3-7t
so v'(t)=derivative of s ---> 15t^2-7
and a'(t)=derivative of v---> 15t
That's what I've understood so far

Answer 12

\[S=5t^3-7t \]
I didn't see the first equation, and thought it was \[s(t) = 15t^2-7\]

Answer 13

However, your notation is incorrect: it's either \[v(t) = \frac{ds}{dt}\] OR \[s'(t) = \frac{ds}{dt}\]\[v'(t) = a(t)\]

Answer 14

\[v'(t) = a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}\]

Answer 15

so \[v(t) = 3*5t^{3-2}-7t^{1-1} = 15t^2-7\]\[a(t) = \frac{dv}{dt} = 2*15t^{2-1} = 30t\]

Answer 16

Ooohhh I didn't see that

Answer 17

Knowing that my acceleration is 30t can I just plug in t=1 ? and for part C is my approach to plug in 0 into v'(t) correct ?

Answer 18

(a) Find velocity & acceleration as functions of t
velocity at time t......... 15t^2-7
acceleration at time t.. ?

Okay, we did (a).

(b) Find the acceleration after 1 second= ? (would I just plug in 1 from the equation in part a, after finding acceleration ?)
Yes, plug in \(t=1\) in our formula for \(a(t)\)
\[a(1) = \]

(c) Find the acceleration at the instant when the velocity is 0 ?
Now we need to solve \(v(t)=0\). With that value of \(t=t_0\), evaluate \(a(t_0) =\)

Answer 19

For part C I got 20.5 ?

Answer 20

Yes, that's correct!

Answer 21

Thank you so much for your help @whpalmer4

Answer 22

You're welcome! as a learning exercise, you should also graph s(t), v(t), and a(t) on the same sheet of graph paper (or on your graphing calculator) and notice how they interact. remember, the first derivative is the slope of the tangent to the function. velocity is the slope of the position function, so when the particle stops moving (to change direction, perhaps), the velocity goes to 0 and then increases again in magnitude as the particle starts moving again.