# Acceleration time graph

1. Mar 12, 2015

### Suraj M

1. The problem statement, all variables and given/known data
A particle starts from rest. Its acceleration vs time graph is as shown. What will be the maximum speed?

2. Relevant equations
usual kinematic equations(3+)
3. The attempt at a solution
velocity at t=11 seconds should be max, so the area under the given graph = change in velocity
So vf-0 = 55
so max. velocity= 55 m/s
But, if i try it this way:
$a=\frac{dv}{dt}$
$a.dt=dv$
integrating..
$a∫_{0}^{11} dt = ∫_{0}^{v} dv$
so $a(11-0)=v-0$
i know this is wrong because $a$ is not constant. But how do i include it?
Any help is appreciated.

Last edited: Mar 12, 2015
2. Mar 12, 2015

### Merlin3189

Since a is a function of time, ∫adt is not a∫dt

You need to write a as a function of t then integrate it correctly.

3. Mar 12, 2015

### Pierce610

But if a particle starts from rest and its acceleration vs time graph is as shown, I ask to myself why at the intial time the initial acceleration is not null?

4. Mar 12, 2015

### Merlin3189

You could equally ask, why the question chose to set time 0 at the moment the acceleration starts.
Seems a sensible choice. Say the acceleration is given by a spring which is compressed. It could sit in that state indefinitely, until you release the particle. Then something interesting happens, so you start measuring from that moment and call that t=0.

To keep you happy, acceleration was 0 or null at t=-1, t=-7.5, t= -0.000000000000000000001, but rises to 10 m/sec2 (or whatever that blurred figure is) at t=0.

5. Mar 12, 2015

### Suraj M

oh yeah, i don't how i missed that..
so $a=10-βt$ where $β= \frac{da}{dt}$
so $β=\frac{10}{11}$
so i get $$∫_{0}^{11}(10 - \frac{10}{11}t)dt =∫_{0}^{v}dv$$
so then i get $10(11-0)-(\frac{10}{11}).(11^2 -0) = v$
so $v = 110-110=0$ huh?

6. Mar 12, 2015

### Staff: Mentor

Try this integration again. Show intermediate steps.

7. Mar 12, 2015

### Suraj M

$$∫_{0}^{11}(10dt -\frac{10}{11}tdt)=∫_{0}^{v_f}dv$$
$$10(t_f -0) - \frac{10}{11}(\frac{t_f^2}{2}-0)=v_f-0$$
so then substituting $t_f=11$
Ok got it.. thanks, sorry for the stupid mistake

Last edited: Mar 12, 2015
8. Mar 12, 2015

### Staff: Mentor

What is the integral of tdt? (Hint: it's not t2)

Chet

9. Mar 12, 2015

### Staff: Mentor

What physical law that you are aware of says that the acceleration of a particle has to be continuous with time?

Chet

10. Mar 12, 2015

### Merlin3189

∫ k dt = kt , but ∫ kt dt ≠ kt2

Try differentiating kt2 and see what you get.

Edit - You posted the right answer while I was writing this!

11. Mar 12, 2015

### Suraj M

Thank you, i corrected it.. in post #7

12. Mar 12, 2015

### Pierce610

Mr. Chet ,
I wanted only say that the particle in question is defined to "start at rest" while in the graph, for how I think to interpretate it,seems to start with an initial acceleration not null. Isn't definition of rest "no acceleration"?

13. Mar 12, 2015

### Staff: Mentor

No. Definition of "at rest" is zero velocity.

Chet

14. Mar 12, 2015

### Pierce610

and then also the acceleration?

15. Mar 12, 2015

### Staff: Mentor

No. Acceleration is rate of change of velocity.

16. Mar 12, 2015

### Pierce610

but then let me know how is the acceleration of a particle at rest, that is with speed null?

17. Mar 12, 2015

### Staff: Mentor

Imagine that, at times t<0, the velocity is zero, and at times t>0, the velocity is kt, where k is a constant. So the velocity is a continuous function of time. What is the acceleration at times t < 0? What is the acceleration at times t > 0? Is the acceleration a continuous function of time at t = 0?

Chet

18. Mar 12, 2015

### Pierce610

I've understand: before a particle moves with constant speed v, it needs to be accelerated.
It can't move from 0 to v instantly. Also in free fall it should be an acceleration not null at the start time, 9.8 m/s^2

19. Mar 12, 2015

### Staff: Mentor

If a stationary particle were to have no acceleration, it would forever remain at rest.