Question

An object moves along the x axis according to the equation
x = 3.50t2 − 2.00t + 3.00,
where x is in meters and t is in seconds.

At what time is the object at rest?

Answer 1

It is at rest when its velocity is equal to zero. So you need to find the velocity, set it to zero and calculate the time from it.

Answer 2

i know that, i just don't know how to calculate it :(

Answer 3

Do you know how to differentiate?

Answer 4

yes, it will be 7t-2

Answer 5

yes, and since you know (well, you should know :D) that the change of x with respect to time is velocity, you just got your velocity :D.
To put it in better words: \[\ v(t)=\frac{dx(t)}{dt} \]

Answer 6

likewise,
the change of velocity with respect to time is acceleration or:\[\ a(t) = \frac{dv(t)}{dt} \]
You don't need this one for this problem but I thought it would be nice to mention it.

Answer 7

i know how to differentiate, i just don't know how to get the time when the object has v=0 (at rest)

Answer 8

ok so you differentiated x(t) with respect to time to get v(t) which is:
v(t) = 7t -2
and you know that it will be at rest when v = 0, so just plug this in and get:
0=7t - 2
t = 2/7 s

Answer 9

yea, i got it now before you posted your answer..it was in front of me the whole time :@

Answer 10

yea :D

Answer 11

can i ask you another question please? :)

Answer 12

sure...

Answer 13

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.4 m/s at ground level. The engines then fire, and the rocket accelerates upward at 3.90 m/s2 until it reaches an altitude of 1130 m. At that point its engines fail, and the rocket goes into free fall, with an acceleration of −9.80 m/s2. (You will need to consider the motion while the engine is operating and the free-fall motion separately.)


What is its velocity just before it hits the ground?

i got 117m/s wrong.

Answer 14

169.2 is also wrong.. :(

Answer 15

1st find what the velocity is at the altitude of 1130m (the rocket is still moving, it's not falling yet!), treat that velocity as your initial velocity. At that point the only acceleration acting on the rocket is gravitational (g=9.81m/s^2).

So now you will have to calculate how much time it takes from that point for a rocket to reach a full stop. once you got that time you can calculate the distance traveled in that time. Add that distance to to altitude of 1130m and you got the highest point the rocket will reach. From here I think you will manage.

Answer 16

okay, the maximum height is 1.83 km

Answer 17

what I mean by "treat that velocity as your initial velocity" is that you treat that velocity as initial velocity for the free-fall motion, not the first one, don't replace it with the 80.4 m/s

Answer 18

i don't get it.

Answer 19

What is the correct answer? I got something so I want to check it

Answer 20

correct answer for maximum height is 1.83 km.
i don't know the answer to the velocity before it hits the ground.

Answer 21

ah... I usually use the principle of conservation of energy here.
Since all potential energy will be converted to kinetic energy just at the point of contact you can write Ep = Ek or : \[\ \frac{mv^2}{2}=mgh\]
Solve this for v and plug in g and h (m cancels out)