Ek physics #59

[berriesandcream]

A particle moving at 10m/s reverses it's direction to move at 10m/s in the opposite direction. If its acceleration is -10m/s^2, what is the total distance that it travels? The answer is 10m.

Okay, so my logic was to use:
V^2=v0^2+2ad

Thing is, don't v and v0have the same magnitude just different direction? So why isn't the answer 0m?

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[ilovemcat]

A particle moving at 10m/s reverses it's direction to move at 10m/s in the opposite direction. If its acceleration is -10m/s^2, what is the total distance that it travels? The answer is 10m.

Okay, so my logic was to use:
V^2=v0^2+2ad

Thing is, don't v and v0have the same magnitude just different direction? So why isn't the answer 0m?

Acceleration is -10 m/s^2.

Initial Velocity is: 10m/s
Final Velocity is: -10m/s

It takes 1 second to go from 10m/s to 0m/s.
It takes 1 second to go from 0m/s to -10m/s.

Average Velocity for the 1st portion of the trip is (5m/s):
(0+10)/2 = 5m/s x 1s = 5m

Average Velocity for the 2nd portion of the trip is (-5m/s):
(0-10)/2 = -5m/s x 1s = -5m

The net displacement is zero.
The total distance is not zero - it's 10m.

 

[ssvector]

In reference to,

"Average Velocity for the 1st portion of the trip is (5m/s):
(0+10)/2 = 5m/s x 1s = 5m"

Where is the "0" coming from in the "(0+10)". And can you expand on what exactly the first portion of the trip is? Thanks

 

[ThirdEye]

In reference to,

"Average Velocity for the 1st portion of the trip is (5m/s):
(0+10)/2 = 5m/s x 1s = 5m"

Where is the "0" coming from in the "(0+10)". And can you expand on what exactly the first portion of the trip is? Thanks

The velocity of 0 is the point when the car changes direction.

Basically the car is doing this:

10m/s ---------> 0 m/s

then this:

-10m/s <--------- 0 m/s (The negative only means diff. direction)

(10 + 0)/2 gives the average velocity of 5m/s.

Since the acceleration (rather deceleration is 10m/s^2), it takes exactly 1 second to change the velocity by 10m/s. (10m/s^2 x 1s = 10m/s)

Take the average velocity for both directions (each way is 5m/s since the numbers are all the same and only direction changed), and multiply by the total time.

5m/s x 2s = 10m

or 5m/s x 1s (going right) + 5m/s x 1s (going left) = 10m

 

[N8D9]

Hi guys, I understand why the total distance is not 0, however I have trouble understanding why we have to break this up into two trips and take the average velocities of each trip (5m/s) to solve the problem?

What I was doing is taking the average velocity of the whole trip (which is obviously 10m/s) and using that to solve the problem (as shown below):

t = (-10m/s - 10m/s)/(-10m/s^2) = 2s -> time taken for the whole trip

distance = speed (10m/s) * time (2 secs)
thus, distance = 20m

I understand what you guys did above but I don't get why this is wrong! (I'm terrible at physics too)

Thanks in advance for your help! : )

 

[advair250]

Your approach is incorrect, since there is a change of direction, you can't think of it in the fashion you have. I'd suggest sitting down, drawing out what is happening with reference to the post above's solution. Simply, you can't use the speed of 10 m/s, as it is not constant. in order to go from + to -, you have to slow down to zero and switch directions, thus you must use the Vavg method outlined.

 

[discowisco]

Solve for both directions separately and add using the equation you used.

if im not mistaken, the kinematic equations work for one direction only.