# Acceleration when only distance is given for a single second

1. Jan 22, 2012

### Izal

Apologizes for being a first time poster. I've been staring at this question to no avail for the last hour, and was looking for some help with it. Any would be appreciated.

1. The problem statement, all variables and given/known data
A body starting from rest travels 100m in the sixth second of its motion. If the acceleration is constant, what is its magnitude?

3. The attempt at a solution
Using the kinematic equations, I have only been able to find the Avg Velocity for the single second, which doesn't help out. Other then that, I'm stumped.

2. Jan 22, 2012

### SHISHKABOB

EDIT WHOOPS I misread the problem entirely disregard this post

3. Jan 22, 2012

### SammyS

Staff Emeritus
Hello Izal. Welcome to PF !

No apologies needed !

This is a bit trickier than I first thought.

From t=5 s to t=6 s the body travels 100m. → You can find the average velocity during this time interval. For constant acceleration, the average velocity is equal to the instantaneous velocity at the middle time of the time interval, t = 5.5 sec. From that you can find the acceleration.

4. Jan 22, 2012

### Izal

The equations I'm using are:
Vav=Δx/Δt
Vav=(Vf+Vi)/2
Vf2=Vi2+2aΔx
Δx=Vit +(1/2)at2
Vf=Vi+at

But with these equations I'm still unsure of how to find acceleration. The only part I've been able to find out the Vav in the 6th second by doing Δx/t giving Vav=100m/s

5. Jan 22, 2012

### Izal

Ha! Alright, so because acceleration is uniform and the Vav is 100m/s over the 6th second then the instantaneous is 100m/s @ t=5.5s which makes a heck of a lot of sence. the answer is then 100m/s=0m/s + a (5.5s) which gives a=18.2m/s2.

Thanks a lot for the help. Appreciate it. It kills me the answer is so easy after spending so long on it.

6. Jan 23, 2012

### Curious3141

I would suggest just using: $s = ut + \frac{1}{2}at^2$

where u = initial velocity, s = distance and t = time to set up two equations.

Here, u = 0 since the boy starts from rest.

Sixth second is from t = 5 to t = 6. So the difference in the displacement at t = 6 and t = 5 is 100m.

The displacement at t = 6 is given by: $s(6) = \frac{1}{2}a*6^2 = 18a$ (equation 1)

The displacement at t = 5 is given by: $s(5) = \frac{1}{2}a*5^2 = 12.5a$ (equation 2).

Take equation 1 - equation 2 and equate the LHS to 100:

100 = 5.5a

a = 100/5.5 = 18.2 m/s^2.

No need to calculate average velocity here (even though it is a perfectly valid method).

Last edited: Jan 23, 2012
7. Jan 23, 2012

### SammyS

Staff Emeritus
Izal,

What you have done here is fine. You're making no more assumptions using average velocity this way than you would be making using the method suggested by Curious3141 and in my opinion, using average velocity the way you did is much simpler.