# Kinematic -- If object in 5th and 6th sec passes 20

1. Jul 20, 2014

### alexstrasuz

1. The problem statement, all variables and given/known data
Object from state of inaction starts to move with equal acceleration. If objec in 5th and 6th sec passes 20, whats acceleration.

2. Relevant equations
s=v0*t+a*t*t/2 v0=0

3. The attempt at a solution
s=a*t*t/2
a=s/2
a=10

2. Jul 20, 2014

### Nathanael

You can't use "v0=0" because at "v0" the object has already been accelerating for 5 seconds.

3. Jul 20, 2014

### Satvik Pandey

Does the object covered 20m distance between 5sec. and 6sec. of the journey?
During the start of journey
u=0.Let the acceleration be a then after 5 sec. the velocity of the object=5a(using v=u+at)
For the motion of object between 5sec. and 6sec.---
u=5a,S=20m(given),t=1sec.
Using S=ut+0.5at^2 Equation
20=5a+0.5a.Find the value of a to get the answer.

4. Jul 21, 2014

### ababystarr

In case you want a quick summary on the topic on Kinematics, here's a good resource:

Last edited by a moderator: Sep 25, 2014
5. Jul 21, 2014

### HallsofIvy

Staff Emeritus
I don't know what you mean by "at v0". At t= 0, the object is just starting to accelerate with initial speed v0= 0 so this formula can be used to find the distances at t= 5 and t= 6.

6. Jul 21, 2014

### Nathanael

By "at v0" I meant the "v0" that the OP was using in his equations. (I wanted to keep it as simple as possible because he didn't seem to be a native english speaker.)

If you look at his equations then you can see he assumed "(Δ)t=1" and "s=20" which means he created his equation to describe the time interval from t=5 to t=6.

That is why I said he can't use v0=0, because HIS "v0" is not 0

(You can certainly solve the problem using v0=0 but my comment was on the method that he attacked the problem with.)