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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?

In both cases acceleration is uniform and initial velocity is zero, so we can double average velocity to get final velocity, and final velocity will be equal to the change in velocity.  Therefore:

On the first ramp

• vAve = `ds/`dt = 10m/8s = 1.25m/s

so

• vf = vAve*2 = 2.5 m/s

so

• aAve = `dv/`dt = 2.5m/s / 8s = .31m/s^2

On the second ramp

• vAve = `ds/`dt = 10m/5s = 2m/s

so

• vf = 2*vAve = 2*2m/s =4m/s

so

• aAve = `dv/`dt = 4m/s / 5s = .8m/s^2

The average rate of change of acceleration with respect to ramp slope is

ave rate = change in acceleration / change in ramp slope

= ( .8 m/s^2 - .31 m/s^2) / ( .10 - .05 )

= (.49 m/s^2) / (.05)

= 9.8 m/s^2,

meaning 9.8 m/s^2 per unit of ramp slope.