Finding acceleration

1. Jan 23, 2015

JaySean

1. Not sure if it's the right forum but I'll try. The velocity of an particle varies quadratic with a formula based on time

a) Find the equation for acceleration and length as a function of time.

b) a = 3 m/s^3 (I think it's alpha)
Find the acceleration when t1=2s t2=5s and how big is the mean acceleration in the time intervall t2-t1.

2. See picture

3. a) I think you can use the formula saying u=0 (start velocity) and say a=(v-u)/t
And do the same for where s= length. But is it that easy or are they looking for a different solution, as this is something I learned few years ago. Or do I have to antiderivate it, which would make it complicated because there is no known fact.

b) If a) is right could I then say solve v(t) which gives me t=3 alpha= 3 --> v(t)=27 m/s, and then get acceleration by the 2nd formula v= 27 u=0 t=3, so acceleration is 9? Or am I on the wrong way?

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2. Jan 23, 2015

Staff: Mentor

That's a bit vague. Is any additional information given? Such as velocity at time 0.

Careful! Those formulas are for constant acceleration, which is not what you have here. (Note how the velocity is linear with time, not quadratic, in the formula.)

3. Jan 23, 2015

JaySean

There is no additional information, but the formula (picture).

Do you have any hint to how I get started on solving b)?

4. Jan 23, 2015

Staff: Mentor

Ah, I see it now. $v(t) = 1/2 \alpha t^2$.

First solve part a).

5. Jan 23, 2015

JaySean

Do I have to solve it as differintial equation? As I'm not allowed to use the formula for constant accerlation?

I know I can derivate v'(t) to get a(t)= at where a= alpha, but how do get to the length formula?

Last edited: Jan 23, 2015
6. Jan 23, 2015

Staff: Mentor

You'll have to use a bit of calculus. (How do distance, velocity, and acceleration relate to each other?)

That's correct. You cannot use the standard constant acceleration formulas here.

7. Jan 23, 2015

JaySean

I know that the distance particle traveled is s(t)= antiderivative of v(t), but wouldn't I need additional information to find the constant?

Edit: Already found the formula for acceleration a(t)= qt where q is a constant

8. Jan 23, 2015

HallsofIvy

Staff Emeritus
Not to find the distance travelled over a given time interval. That is the difference between two lengths

9. Jan 23, 2015

Staff: Mentor

To express the position, you'd need additional info. But for the distance traveled, you can just set s(t=0) = 0.

Good. (That constant is given as $\alpha$.)

10. Jan 24, 2015

JaySean

(I didn't know how to write alpha so I just called it q)
So the formula for poisition as a function of time is S(t)=0 when t=0?

and for b) since a(t)=qt where q=3 m/s^3 for t1= 2s and t2=5s
a(2)=6 m/s^2 and a(5)=15 m/s^2
mean accleration= 11/3 m/s^2

11. Jan 24, 2015

Staff: Mentor

Yes, that will give you the position from the starting point (where t = 0, presumably).

Good.

How did you calculate that?

12. Jan 24, 2015

JaySean

Sorry I meant 21/3 m/s^2 which is 7m/s^2
a(2)+a(5) =6+15=21
And t2-t1=5-2=3

13. Jan 24, 2015

Staff: Mentor

Careful! What's the definition of average acceleration?

14. Jan 24, 2015

JaySean

Since I already know what alpha is and time is I might be able to use the general formula for acceleration knowing the velocity at beginning and end.
This gives me
vo=3*(2)^2=12
v1=3*(5)^2=75

acceleration= (75-12)/(5-2)=21

15. Jan 24, 2015

Staff: Mentor

Good.

You left out the factor of 1/2.

16. Jan 24, 2015

JaySean

Sorry this is stupid of me,
(37,5-6)/3 =10,5
v1-v0/delta(t)

I have another question not sure if anyone is able to answer:

Car with velocity 36 km/h colliding and stoping over a length of 1 meter (end velocity = 0). Find acceleration.
I have solved it the simple way using v^2 = u^2 + 2as and got the right answer.
Now my question is how do I set it up as an differential equation.

My attempt:
Since I know S(0)= 1 (saying start position is 1 and end is 0)
velocity start = 36km/h and end is 0
I also know that position is antiderivative of velocity
S(t)=∫36
S(t)= 36t+C

Attempt on finding constant c.
S(0)=36*0+C=1
C= 1
S(t)= 36t+1

17. Jan 24, 2015

Staff: Mentor

I would say that the start position is 0 and the end position is 1.

You cannot assume that the velocity is constant. (After all, you are told that it is stopping.)

All you'll do, by setting up a differential equation, is derive those kinematic formulas. Nothing wrong with that, as long as you start with the correct assumptions. Assume that the acceleration is constant, then you can integrate to get the position as a function of time.

18. Jan 25, 2015

JaySean

Do you mean derivating for example the formula
v=v0+at?

19. Jan 25, 2015

Staff: Mentor

Yes. And, using a few tricks, all the others.