# Thread: why don't the t's cancel?

1. ## why don't the t's cancel?

Hi,

This should probably go in the physics help forum website, but phf seems to have a few glitches that have made it difficult for me to post there.

Anyways, it's really just a simple algebra question that has probably been asked thousands of times:

In Newton's distance formula for a constantly accelerating object, to wit:

$d=v_it+\frac{at^2}{2}$

if $a=(\frac{v_f-v_i}{t})$

so that

$d=v_it+\frac{t^2}{2}(\frac{v_f-v_i}{t})$

why don't the t's cancel??

I.e. why isn't Newton's formula instead:

$d=v_it+\frac{t(v_f-v_i)}{2}$

2. Originally Posted by rainer
Hi,

This should probably go in the physics help forum website, but phf seems to have a few glitches that have made it difficult for me to post there.

Anyways, it's really just a simple algebra question that has probably been asked thousands of times:

In Newton's distance formula for a constantly accelerating object, to wit:

$d=v_it+\frac{at^2}{2}$

if $a=(\frac{v_f-v_i}{t})$

so that

$d=v_it+\frac{t^2}{2}(\frac{v_f-v_i}{t})$

why don't the t's cancel??

I.e. why isn't Newton's formula instead:

$d=v_it+\frac{t(v_f-v_i)}{2}$
HI

They should cancel each other . WHy not ?

3. They do cancel, it's just that the acceleration can't always be expressed like that, i.e. $a'(t) \neq const$

4. Just as the others mentioned already. Most of the time, you will need the formula with both the t's to solve the question. It really depends on the type of question and values given. Then you can mold the formula accordingly.

5. Hello rainer
Originally Posted by rainer
Hi,

This should probably go in the physics help forum website, but phf seems to have a few glitches that have made it difficult for me to post there.

Anyways, it's really just a simple algebra question that has probably been asked thousands of times:

In Newton's distance formula for a constantly accelerating object, to wit:

$d=v_it+\frac{at^2}{2}$

if $a=(\frac{v_f-v_i}{t})$

so that

$d=v_it+\frac{t^2}{2}(\frac{v_f-v_i}{t})$

why don't the t's cancel??

I.e. why isn't Newton's formula instead:

$d=v_it+\frac{t(v_f-v_i)}{2}$
They do cancel, and the formula you have derived is correct. It can then be simplified to:

$d = \tfrac12(v_i+v_f)t$

which simply says that the distance travelled is equal to the average of the initial and final velocities multiplied by the time taken. Which, for constant acceleration, is correct.

(Geometrically, the distance travelled = area under the v-t graph. For constant acceleration this is a straight line graph, and the formula is simply the area of a trapezium.)