# Determine the times at which the velocity=0

1. Jan 3, 2006

### teng125

the acce of a particle is a=A - 6t^2 where A is a constant. At t=0,the particle starts at x=8m with v=0. Knowing that at t=1s, v=30m/s.
Determine the times at which the velocity=0

pls help...
the answer is 0 and 4s

2. Jan 3, 2006

### Galileo

Now that you are given the acceleration of the particle, do you know how to find the velocity from the initial condition? (Remember: acceleration is the time derivative of the velocity).

3. Jan 4, 2006

### teng125

ya,i tried to get dv/dt=A-6t^2 to v-v(initial)=(At-2t^3) - (At[initial] - 2t^3[initial]).but i think it is wrong....
what eqn do u get to form??

4. Jan 4, 2006

### AlphaNumeric

$$v = At - 2t^{3} + C$$

You know v(0)=0 and v(1) = 30, so you can work out A and C. Once you have that, solve v = 0.

5. Jan 4, 2006

### teng125

for part b) what is the total distance travelled by the particle when t=5s??

i try to integrate and get x=16t^2 - 1/2t^4 + 8 but when i try to do,the answer is wrong.

6. Jan 4, 2006

### Fermat

The particle first moves away from the origin until v = 0.
How far has it travelled ? How long did it take ?

How much further does it travel, on the way back, until t = 5 sec ?

7. Jan 4, 2006

### teng125

firstly is my eqn correct above??
i try to find all the way from t=1 to 5 and found it goes back...what should i do

8. Jan 4, 2006

### Fermat

You just have two distances, the distance out and then a bit more, coming back, to find.

Think of it like a particle in shm. It starts at the origin. Moves out a distance and then moves back towards the origin again.
That is what this particle is doing, although the movement is certainly not shm.

Find out when the velocity = 0. That is when the particle is at its max. distance from the origin. What is the value of t ?
Use your eqn to find out the distance, x, from the origin. That is how far it has travelled so far.
Now find its distance from the origin at t = 5. What distance has the particle moved back by ?
Now what is the total distance moved ?

9. Jan 4, 2006

### teng125

i thought the problem stated that at t=0,x=8 and v=0 also.so it started at 8 for t=0.
then at t=5s, i got 95.5

is it correct??

10. Jan 4, 2006

### mukundpa

x in the equation is displacement, not the distance travelled. If you move 100 m and come back adistance 20 m, the total distance travelled is 120 m but the displacement is only 80 m. In the same way the particle is moving forward till v=0 and then comes back, you have to find distance travelled

11. Jan 4, 2006

### Fermat

Everything in your last post is correct, but ...

The particle starts from the origin, moves out by about 130 odd metres, stops momentarily (v=0, t < 5) then moves back again towards the origin such that at t = 5, its distance from the origin is then 95.5 m
You have to (first of all) find out its distance from the origin, when v=0.

What's the eqn for velocity ?

12. Jan 6, 2006

### teng125

now when v=0,t=- or + 4s.so i substituted into the eqn x=16t^2 - 1/2t^4 + 8 , i got 120.
then i try to take 120-95.5=24.5 and then plus 120 =144.5.However,the answer is incorrect.Where did i go wrong??

13. Jan 6, 2006