# Confusing Physics Question

• February 23rd 2013, 09:20 AM
sakonpure6
Confusing Physics Question
1-Fat Albert is standing by the classroom door at 11:40. Only 0.30 s after the bell rings he is running at 4.0m/s towards the cafeteria.

a) What was his average acceleration during this time?

b)How far did he move during this time?

First of all, how can we get acceleration if we are dealing with scalar values? and for this part of the question "only 0.30s after the bell" do we assume he was still for the beginning of those .30 s? My answer would be a=0 because he would be running at a constant speed of 4.0m/s . Am I right?

For part b) how can we calculate this without the total time it took him to reach the cafeteria?!

Any help is appreciated ! Thank you in advance!
• February 23rd 2013, 09:31 AM
ILikeSerena
Re: Confusing Physics Question
Quote:

Originally Posted by sakonpure6
1-Fat Albert is standing by the classroom door at 11:40. Only 0.30 s after the bell rings he is running at 4.0m/s towards the cafeteria.

a) What was his average acceleration during this time?

b)How far did he move during this time?

First of all, how can we get acceleration if we are dealing with scalar values? and for this part of the question "only 0.30s after the bell" do we assume he was still for the beginning of those .30 s? My answer would be a=0 because he would be running at a constant speed of 4.0m/s . Am I right?

For part b) how can we calculate this without the total time it took him to reach the cafeteria?!

Any help is appreciated ! Thank you in advance!

Hi sakonpure6! :)

The word acceleration can mean both a vector or a scalar.
In this problem we'd expect a linear movement, which means it does not matter.

Average acceleration is:

$a_{average} = \frac{\Delta v}{\Delta t}$

What was his velocity when the bell rang?

For part (b), suppose we treat the acceleration as constant, starting from the time the bell rang, how far did Albert get then?
• February 23rd 2013, 09:38 AM
sakonpure6
Re: Confusing Physics Question
So for part b we would use this formula d= v1(t)+ 1/2(a)(t)^2 which results a movement of about 1.8 m! thank you!
• February 23rd 2013, 10:13 AM
ILikeSerena
Re: Confusing Physics Question
Quote:

Originally Posted by sakonpure6
So for part b we would use this formula d= v1(t)+ 1/2(a)(t)^2 which results a movement of about 1.8 m! thank you!

Not quite. Typo?

Did you use:
v1=0
t=0.3
a=4 / 0.3
• February 23rd 2013, 10:20 AM
sakonpure6
Re: Confusing Physics Question
Quote:

Originally Posted by ILikeSerena
Not quite. Typo?

Did you use:
v1=0
t=0.3
a=4 / 0.3

yea v=4 m/s
and t= 0.30

• February 23rd 2013, 10:27 AM
ILikeSerena
Re: Confusing Physics Question
Quote:

Originally Posted by sakonpure6
yea v=4 m/s
and t= 0.30

$d = 0 \cdot 0.3 + \frac 1 2 \cdot \frac{4}{0.3} \cdot (0.3)^2 = ?$