Kinematics

[chiddler]

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EK #61

Particle moving at 10 m/s reverses to move at 20 in opposite direction. If a = -10 m/s^2, what is total distance that it travels.

Answer is 25.

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Here is my question:

Vf^2 = Vi^2 + 2ad
-400 = 100 + 2(-10)(d)
d = 25

but specifically, i'm looking at the negative sign. Isn't -20^2 = 400 positive? How is my thinking wrong? When should the velocities be negative?

Many thanks.

 

[MedPR]

EK #61

Particle moving at 10 m/s reverses to move at 20 in opposite direction. If a = -10 m/s^2, what is total distance that it travels.

Answer is 25.

--

Here is my question:

Vf^2 = Vi^2 + 2ad
-400 = 100 + 2(-10)(d)
d = 25

but specifically, i'm looking at the negative sign. Isn't -20^2 = 400 positive? How is my thinking wrong? When should the velocities be negative?

Many thanks.

Since you're solving for distance, you should be thinking of V as speed instead of velocity. Distance and speed are scalars, and therefore are always positive. Velocities are negative when you are comparing two different velocities in opposite directions (If you designate North as positive,then someone traveling 100m/s South is traveling with a velocity of -100m/s).

 

[chiddler]

Since you're solving for distance, you should be thinking of V as speed instead of velocity. Distance and speed are scalars, and therefore are always positive. Velocities are negative when you are comparing two different velocities in opposite directions (If you designate North as positive,then someone traveling 100m/s South is traveling with a velocity of -100m/s).

but if i think of it as speed (absolute value of velocity), then it becomes

400 = 100 + 2(10)d

which gives me the wrong answer

 

[MedPR]

but if i think of it as speed (absolute value of velocity), then it becomes

400 = 100 + 2(10)d

which gives me the wrong answer

You're forgetting that the particle is changing directions, so it's not the same as if you increased speed from 10 to 20 in the same direction.

x=1/2at^2 ----> x=5 (vf-vo/a)^2

For decelerating from 10 to 0m/s
x = -5

For accelerating from 0 to 20
x=-20

total distance = 25m

Consider the initial direction to be north, and the final direction to be south. The acceleration is always south, so you leave the sign the same for both legs of the trip (10m/s to 0, and 0m/s to 20). I chose to make my acceleration negative, but if you set it to +10m/s^2 you still get 25m.

 

[chiddler]

okie dokie well understood. thanks for the help