A simple question...

[Temperature101]

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How do we know when to use Velocity = Distance/Time versus one of the other four kinematics equations? Are there any operative words that give you that clue?

 

[StarlingForce1]

Depends what data you are given, what you need to find, and what equation/combination of equations allow you to accomplish that.

 

[Temperature101]

Depends what data you are given, what you need to find, and what equation/combination of equations allow you to accomplish that.

Here is an example

A car moving at 20 m/s brakes and slides to a stop. If the coefficient of kinetic friction between the pavement and the tires of the car is .1, how far does the car slide? how much time is needed for the car to come to a stop?

Choices for first part of the question:
a) 50m
b) 100m
c) 200m
d) 400m

Second part:
a) 1s
b) 10s
c) 20s
d) 40s


ma = N u
ma = mg u (cancel m out)
a = g u
a = 10 (0.1) = 1

Then, solve for x from V^2 = Vo^2 + 2ax
0 = (20)^2 + 2(1)x
-400/2 = x
x = 200

Then for time
V = Vo + at
0 = 20 + 1t
t = 20

If I use time = distance/velocity, I will find a different answer.... Why I cant use it in that case?

 

[bpatient]

I would like to know, also where did you get this question from? TBR?

 

[N1st]

Here is an example

A car moving at 20 m/s brakes and slides to a stop. If the coefficient of kinetic friction between the pavement and the tires of the car is .1, how far does the car slide? how much time is needed for the car to come to a stop?

Choices for first part of the question:
a) 50m
b) 100m
c) 200m
d) 400m

Second part:
a) 1s
b) 10s
c) 20s
d) 40s


ma = N u
ma = mg u (cancel m out)
a = g u
a = 10 (0.1) = 1

Then, solve for x from V^2 = Vo^2 + 2ax
0 = (20)^2 + 2(1)x
-400/2 = x
x = 200

Then for time
V = Vo + at
0 = 20 + 1t
t = 20

If I use time = distance/velocity, I will find a different answer.... Why I cant use it in that case?

Because the velocity is changing. You could use the average velocity though (so long as you have a constant acceleration as you do in this case).

For a constant acceleration, v-average = (vinitial + vfinal) / 2. So for this question, the average velocity is 0 + 20 / 2 = 10. 200 / 10 makes 20 seconds.

v = d/t, so d= v * t, is only true if the the value of v applies to the whole time that you are talking about.

 

[Temperature101]

Because the velocity is changing. You could use the average velocity though (so long as you have a constant acceleration as you do in this case).

For a constant acceleration, v-average = (vinitial + vfinal) / 2. So for this question, the average velocity is 0 + 20 / 2 = 10. 200 / 10 makes 20 seconds.

v = d/t, so d= v * t, is only true if the the value of v applies to the whole time that you are talking about.

Thanks...I have to remind myself constant acceleration means velocity is changing.

 

[Temperature101]

i would like to know, also where did you get this question from? Tbr?

ek 1001