Seed Question 72

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:

This is asking for an average rate. This is a change in something over a change in something. I am being asked to find the average change rate of the acceleration. Here is what I am to find:

Average rate of automobile acceleration = change in acceleration / change in slope of incline

First I find the acceleration of both instances(the slopes are given.)

First one:

I need to find the final velocity, I do this by finding the average first.

‘ds = vAve * ‘dt

10 m = vAve * 8

vAve = 1.25 m /sec

vAve = (v0 + vf) / 2

1.25 = vf/2 (v0 is 0)

Vf = 2.5 m /sec

Now I find the acceleration

Vf^2 = v0^2 + 2a’ds

(2.5)^2 = 2a(10

6.25 = 20a

A1 = 0.3125 m /sec^2

Now I find the acceleration of the second instance doing the same procedure.

‘ds = vAve * ‘dt

10 m = vAve * 5

2 m /sec = vAve

vAve = (v0+vf) / 2

2 m/sec = vf/2

Vf = 4 m /sec

Vf^2 = v0^2 + 2a’ds

(4)^2 = 2a(10)

16 = 20a

A2 = 0.8 m /sec^2

Now I plug in my values to find the average change in acceleration over change in slope.

Change in acceleration / change in slope

(0.8 m /sec^2 – 0.3125 m /sec^2) / (0.1in. – 0.05in.

= (0.4875m/sec^2) / 0.05 in)

Average rate = 9.75 m /sec^2 / in.

The slope isn't in units of in. The slope is a pure number (if rise and run are measured in the same units then rise/run is unitless).

Otherwise excellent. The requested rate is just plain 9.75 m/s^2.