A construction crew uprights an overturned backhoe on a steep hill, whereupon it

Question

A construction crew uprights an overturned backhoe on a steep hill, whereupon it promptly rolls down the hill. (See video "help" button at the bottom of this question to understand the situation better.) Let's suppose the backhoe picks up speed until it reaches level ground and starts slowing down, so that its velocity vs time is as shown m the graph below. Answer the questions following the graph. Assume that all numbers from the graph are good to plusminus 2%. Assume that the deceleration continues past the edge of the graph until the backhoe stops. What is the backhoe s average velocity m the first 3 s? What is the backhoe s average acceleration between t = 0 s and t=3 s? What is the backhoe s average acceleration after it reaches the level ground? How far away is the level ground from where the backhoe started? How far does the backhoe travel on the level ground before coining to a stop?

Explanation / Answer

As seen from the graph the backhoe's average velocity in the first 3 sec=11m/s

Acceleration= Rate of change in velocity

velocity at t=0 is 0

velcity at t=3 is is11m/s

therefore acceleration=11-0/3-0

=11/3

=3.6m/s2

Since when the backhoe is on a steep hill its speed increases abruptly and then when it reaches the level ground its speed starts decreasing

This speed when it reaches level ground=11m/s

acceleation=speed/time

=11/3=3.6m/s2

distance=speed * time

Speed at which backhoe started=11m/s

time at which backhoes started=3s

Distance=11*3=33m

So level ground is 33 m from where backhoe started

Backhoe stopped after reaching speed=6m/s

backhoe stopped after time=12s

distance travelled by backhoe before stopping=speed*time

=6*12=72m

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