<em><u>1.car</u></em><em><u> </u></em><em><u>towing</u></em>
<em><u>2.pulling</u></em><em><u> </u></em><em><u>bucket</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>water</u></em>
<em><u>3.gym</u></em><em><u> </u></em><em><u>equipment</u></em><em><u> </u></em>
<em><u>4.crane</u></em><em><u> </u></em><em><u>machine</u></em>
<em><u>5.tug</u></em><em><u> </u></em><em><u>of</u></em><em><u> </u></em><em><u>war</u></em>
<span>the classification system</span>
<span>motion of truck constitutes of 3 travels.
1. accelerating uniformly with acceleration a1 = 2 m/s^2 until its velocity reached 20 m/s travelling a
distance of 's1' meters.
2. uniform motion with 20 m/s for a time duration t1 = 20s travelling a distance of 's2' meters.
3. uniform deceleration for t2 = 5 sec which stops the truck after travelling a distance of 's3' meters.. </span>
Answer:
W = y (b-a) / ab
Explanation:
Work is defined by the expression
W = ∫ F. dr
In this case the force is in the same direction of displacement, so the scalar product is reduced to the ordinary product
W = ∫ F dr
The expression of the strength left is
F = -y / x²
let's replace and integrate
W = ∫ (-y / x²) dx
W = -y (-1 / x)
We evaluate between the lower limit x = b + a to the upper limit x = 0 + a
W = -y (-1 / b + 1 / a)
W = y (b-a) / ab
where (b-a) is the distance traveled
