0
$\begingroup$

Let's consider a simple school problem.

A car starts moving during 3 seconds with a constant acceleration of 1 m/s^2. Then it stops accelerating and moves 3 seconds more with a constant speed. Find the average acceleration of the car.

Option 1

Calculating the acceleration as the change of velocity divided by the change of time.

Option 2

Using the formula for the uniform accelerated motion.

Why am not getting the same response?

$\endgroup$
2
$\begingroup$

The first equation holds good for average acceleration, but the second is the equation for uniform acceleration. The value obtained using option 1 is correct.

In the time interval from 0 to 6 s, the acceleration changes (a constant value from 0 to 3 s and another constant value from 3 to 6 s). Then you cannot apply the uniform acceleration equation as given in option 2 directly. Find the average distance traveled from 0 to 3 s (where the acceleration is uniform) and then substitute it in the equation. Now, again do this for 3 to 6 seconds (where the acceleration is zero). Hence the average acceleration will be the average of the two calculated accelerations.

That is find $a_1$ for 0 to 3 s using the second equation (I don't know why you do this because it is already given) and then find $a_2$ for 3 to 6 s. The average acceleration will be the average of both $a_1$ and $a_2$

$\endgroup$
0
$\begingroup$

Average acceleration is defined the same way as average velocity :

Average velocity is change in displacement / change in time.
Average acceleration is change in velocity / change in time.

Your 1st calculation gives the constant acceleration which would give the same change in velocity in the same time. This is correct.
Your 2nd calculation gives the constant acceleration which would give the same change in displacement in the same time. This is incorrect.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.