Question regarding kinematics : motion in one dimension.

1. Jun 8, 2012

sankalpmittal

1. The problem statement, all variables and given/known data

The velocity of particle is 0 at t=0. Then ,

(a) The acceleration at t=0 must be zero.

(b) The acceleration at t=0 may be zero.

(c) If the acceleration is 0 from t=0 to t=10 s , the speed is also zero in this interval.

(d) If the speed is 0 from t=0 to t=10 s , the acceleration is also zero in this interval.

2. Relevant equations

All the kinematic equation. (Do you want me to list all of 'em !)

3. The attempt at a solution

At an instant t=0 , the instantaneous acceleration also called acceleration will also be zero. If velocity - time graph is a curve , then velocity will be a function of time and then when we differentiate it w.r.t time we get acceleration at an instant. Putting value t=0 there , I get acceleration=0

So correct options are :
(a) , (c) and (d).

Huh ! This is not the answer given in my textbook. It says (b) , (c) and (d) !

Where did I go wrong ?

2. Jun 8, 2012

Infinitum

Say the time t=0 was defined when a ball thrown up and is just starting its motion downwards again, at the instant v=0. Is the acceleration 0 then?

3. Jun 8, 2012

ehild

You drop a stone at t=0. Its initial velocity is zero, but what about the acceleration?

You hold the stone in your hand at t=0 and you don not release it. What is its acceleration?

Think of uniform circular motion.

ehild

4. Jun 8, 2012

azizlwl

The velocity of particle is 0 at t=0. Then ,

(a) The acceleration at t=0 must be zero.
(b) The acceleration at t=0 may be zero.
(c) If the acceleration is 0 from t=0 to t=10 s , the speed is also zero in this interval.
(d) If the speed is 0 from t=0 to t=10 s , the acceleration is also zero in this interval.

...........
(a) No.At time Δt, the velocity might not be zero. So the possibilty of a≠0
(b) Yes, as above. Gradient may be zero or not.
(c)Yes. No changes of velocity from zero, thus not magnitude and direction change. Speed is zero since magnitude remains zero.
(d)Yes. Magnitude remains zero, velocity constant. Acceleration is zero.

Last edited: Jun 8, 2012
5. Jun 8, 2012

sankalpmittal

No , the acceleration is downward. I know this because I pondered over this question a lot after attempting. But what is confusing me is that mathematically I fail to analyze. If I draw a velocity - time , graph a curve , and differentiate velocity w.r.t time , I get acceleration as a function of time. When I put t=0 , I get a=0.

Acceleration downwards = -9.8 m/s2

a=0

Please consider one dimensional motion only. Thanks.

I fail to get this mathematically. Please see "The attempt at a solution" in my first post.
Are you saying that if Δt is infinitely small , then very very near t=0 , there is possibility that v≠0 ?

Also considering a local maxima and local minima at a point in a displacement-time graph ( a curve ) I get velocity of the body 0 ( tangent parallel to X axis ). When I differentiate a function of time ( displacement ) I get velocity = 0. When I again differentiate it with respect to time I get acceleration as a function of time. When I put t = 0, again I get acceleration = 0 but theoretically in former case, I get acceleration less than 0 while in latter I get it greater than 0.

Last edited: Jun 8, 2012
6. Jun 8, 2012

I like Serena

Hey Sankalp! :)

Let's drop a stone:
$$y(t)=y_0- \frac 1 2 g t^2$$

Differentiate to get:
$$v(t)=-gt$$
So at time t=0, you have indeed v(0)=0.

Differentiate again to get:
$$a(t)=-g$$
Note that acceleration at time t=0 is not zero.

7. Jun 9, 2012

sankalpmittal

Hi ILS ! :)

Suppose I have a velocity-time graph which is a curve and at point t=0 let the velocity(v) of the body be 0.
And velocity is a function of time.

Let
v= t + t2 + t3 + 89
If I differentiate it w.r.t to time , I get acceleration at an instant

dv/dt = 1 + 2t + 3t2
a= 1 + 2t + 3t2
putting t=0
a=1 !

If I have displacement time graph a curve , and say at t=0 ,displacement(s)=0 and displacement is a function of time say :

s= t + t2 + t3 + 89
ds/dt = 1 + 2t + 3t2
d/dt(ds/dt)= 6t + 2
a= 6t + 2
putting t=0 I get a=2 !

Just one more question though after getting this mathematically :

We are measuring instantaneous velocity and acceleration of body at very very near point t=0 and not at t=0 (because it will give indeterminate) such that at t=t0 I have t0- 0 = Δt and Δt→0. Am I right ?

8. Jun 9, 2012

Infinitum

Yep!

Velocity cannot be defined without a time interval, so for instantaneous you just have the interval tending to zero. So basically, you're looking at the limit,

$$\lim_{\Delta t \to 0} \frac{\Delta x}{\Delta t} = \frac{dx}{dt}$$

9. Jun 9, 2012

ehild

If a certain motion started at t=0 with initial velocity v(0), and you want to find the instantaneous acceleration at t =0 you need to measure the velocity at later instants, t=0+Δt. Of course, you can not measure before the start, but the acceleration at zero is defined as the limit Δv/Δt = (v(t)-v(0))/Δt when Δt →0 through positive values.
If you plot v(t) in terms of time, you can draw a tangent to the v(t) curve at t=0. Its slope is the acceleration at t=0. That is true either v(0)= 0 or not.

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10. Jun 9, 2012

sankalpmittal

Thanks ehild , Infinitum , ILS and azizlwl for your excellent replies !

That's were the replies I was expecting !

Now I get it theoretically , graphically and mathematically as well.

Thanks again.