Ladybug Motion

#$&*

course PHY 241

11/18 8

An automobile rolls 10 meters down a constant incline with slope .05 in 8 seconds, starting from rest. The same automobile requires 5 seconds to roll the same distance down an incline with slope .10, again starting from rest.

At what average rate is the automobile's acceleration changing with respect to the slope of the incline?

answer/question/discussion:

vAve_1 = 10 / 8 = 1.25 m / s = v_f / 2

v_f = 2.5 m/s

a = 2.5 / 8 = .3 m/s^2

vAve_2 = 10 / 5 = 2 m/s = v_f / 2

v_f = 4 m/s

a = 4 / 5 = .8 m/s^2

rate of change of a with respect to m = `da / `dm = .5 / .05 = 10 m/s^2 / unit of slope

#$&*"

Ladybug Motion

#$&*

course PHY 241

11/18 8

An automobile rolls 10 meters down a constant incline with slope .05 in 8 seconds, starting from rest. The same automobile requires 5 seconds to roll the same distance down an incline with slope .10, again starting from rest.

At what average rate is the automobile's acceleration changing with respect to the slope of the incline?

answer/question/discussion:

vAve_1 = 10 / 8 = 1.25 m / s = v_f / 2

v_f = 2.5 m/s

a = 2.5 / 8 = .3 m/s^2

vAve_2 = 10 / 5 = 2 m/s = v_f / 2

v_f = 4 m/s

a = 4 / 5 = .8 m/s^2

rate of change of a with respect to m = `da / `dm = .5 / .05 = 10 m/s^2 / unit of slope

#$&*"

#*&!

Ladybug Motion

#$&*

course PHY 241

11/18 8

An automobile rolls 10 meters down a constant incline with slope .05 in 8 seconds, starting from rest. The same automobile requires 5 seconds to roll the same distance down an incline with slope .10, again starting from rest.

At what average rate is the automobile's acceleration changing with respect to the slope of the incline?

answer/question/discussion:

vAve_1 = 10 / 8 = 1.25 m / s = v_f / 2

v_f = 2.5 m/s

a = 2.5 / 8 = .3 m/s^2

vAve_2 = 10 / 5 = 2 m/s = v_f / 2

v_f = 4 m/s

a = 4 / 5 = .8 m/s^2

rate of change of a with respect to m = `da / `dm = .5 / .05 = 10 m/s^2 / unit of slope

#$&*"

#*&!#*&!

&#This looks very good. Let me know if you have any questions. &#