How Do You Calculate Average Speed with Variable Velocities and Accelerations?

[Haniszmi]

Homework Statement

A particle that is moving along a straight line decelerates uniformly from 0.4 m/s to 0.2 m/s in 5 seconds and then has a constant acceleration of 0.2 m/s^2 during the next 4 seconds. What is the average speed over the whole time interval?

Homework Equations

vavg = (x - xo)/t

The Attempt at a Solution

I first broke up the problem into two parts, first part being..."A particle that is moving along a straight line decelerates uniformly from 0.4 m/s to 0.2 m/s in 5 seconds"...I found the deceleration to be -.04m/s^2 by simple plugging into the formula. For the second part..."then has a constant acceleration of 0.2 m/s^2 during the next 4 seconds. What is the average speed over the whole time interval?" Since the acceleration was given to me in that statement I just added the deceleration and acceleration together and divided by two to get the average acceleration. DID I DO THIS CORRECTLY? MY ANSWER WAS .08m/s^2

 

[fifeplayer]

You have the correct relevant equation -
vavg = (x - xo)/t

but in order to find the average speed, you need to find the distance traveled (which you can get from the accelerations and times given), and then divide the distance by the total time.

You can tell your original answer is wrong by checking the units of it - speed is m/s, acceleration is m/s^2 which are the units that adding the accelerations together would give you.

 

[Haniszmi]

Ok I see where I had made my mistake, but now I am confused with how to find the distance traveled. How do you find the distance traveled if there are 2 different velocities?

 

[LowlyPion]

Ok I see where I had made my mistake, but now I am confused with how to find the distance traveled. How do you find the distance traveled if there are 2 different velocities?

In the deceleration phase you can determine that from the relation that

Vf - Vi = a*t

Then with that acceleration number you can figure distance with

Vf^2 - Vi^2 = 2 a (xf-xi)