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!

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:

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

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?

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!

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

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

so a= about 13 m/s^2

Re: Confusing Physics Question

Quote:

Originally Posted by

**sakonpure6** yea v=4 m/s

and t= 0.30

so a= about 13 m/s^2

Yes.

And:

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

Re: Confusing Physics Question

= 0.6 m !!! I used v2 instead of v1! thank you!