quick question -- acceleration when v=0 (1 Viewer)

[blackops23]

Let's say v = 40 - 10 t

so v = 0, when t=4 (i.e. particle is stationary at t=4)

What is the acceleration at t=4 (i.e. what is the acceleration when v=0 (particle is stationary))?

Is it just -10m/s/s?? or is it like some undefined crap?

Thanks, appreciate the help!

EDIT:

Also, say the particle moves according to the equation x=5t(8-t)

How do you find the distance traveled in the first 12 seconds? Could someone please show me?

Thanks

 

[funnytomato]

Let's say v = 40 - 10 t

so v = 0, when t=4 (i.e. particle is stationary at t=4)

What is the acceleration at t=4 (i.e. what is the acceleration when v=0 (particle is stationary))?

Is it just -10m/s/s?? or is it like some undefined crap?

Thanks, appreciate the help!

yes

 

[funnytomato]

the example you've mentioned is like a projectile where you have some initial +ve velocity, but there's a constant -ve acceleration

so it travels in the +ve direction for some time, then velocity decrease to 0 at a certain time(where there is stlll a constant NEGATIVE acceleration), then velocity becomes more and more negative, which mean it's falling back due to this acceleration

mathematically, v=40-10t
a=dv/dt=-10, so there is ALWAYS a negative acceleration
x=integrate v with respect to t = 40t-5t^2 , where x is the displacement(http://www.wolframalpha.com/input/?i=plot+40t-5t^2)

so at t=4, velocity is 0, displacement is 80
but due to the -ve acceleration, it must fall back again

 

[funnytomato]

EDIT:

Also, say the particle moves according to the equation x=5t(8-t)

How do you find the distance traveled in the first 12 seconds? Could someone please show me?

Thanks

distance travelled
after graphing 5t(8-t) http://www.wolframalpha.com/input/?i=plot+5t(8-t)
u can see that it reaches a maximum positive displacement of 80 (use some calculus or symmetry of parabola)
then it comes back to 0 at time 8, having a -ve displacement of 80
then it goes into the -ve direction, at time 12, displacement=-240, distance=240

total distance= 80(+ve displacement for 0-4 seconds) + 80(-ve disp. for 4- 8 seconds) + 240(-ve displacement for 8-12 seconds)=400

 

[blackops23]

Thankyou, I also got 400, just a question, I'm familiar with terms acceleration, velocity and displacement.

But what is this +ve, -ve (acceleration, velocity, displacement) business? I don't know what it is? Could someone please explain?

Thank you.

 

[funnytomato]

Thankyou, I also got 400, just a question, I'm familiar with terms acceleration, velocity and displacement.

But what is this +ve, -ve (acceleration, velocity, displacement) business? I don't know what it is? Could someone please explain?

Thank you.

let's call the displacment x, velocity v, acceleration a

displacement is like distance , except for it's a vector. i.e. it's got direction
for example, if u travel due east for 20m , then west 20m your displacement would be 0 cos you're back to where you've started
but the actual distance you've covered is 40m
if u travel 40m west instead of 20m, ur displacement would be 20m to the west, with distance=60m

so velocity means how fast and which direction you're going, it's the rate of change of your displacement with respect to time
v= dx/dt
for example if v is positive, displacement is increasing with respect to time
"speed" is the magnitude of the velocity, it's like distance in the sense that they don't have direction

acceleration is how fast and which direction ur velocity is changing with respect to time, it's the rate of change of velocity with respect to time
a=dv/dt
say you've got a positive velocity , but the acceleration is negative
that means your velocity is decreasing (going to the -ve direction, not just decreasing speed)