There is no need to assume the belts are of equal length. Lets say PQ = x and QR = y. We know the belts move in opposite direction that is towards Q.
P-----------x------------->Q<--------------y---------------R.
Now the cart travels from P to Q and Q to R, then again from R to Q and finally from Q to P.
Speed of the belt = 0.5 m/s,
Speed of the cart = 2 m/s.
Speed when cart moves from P->Q = 2.5m/s,
Q->R = 1.5m/s,
R->Q = 2.5m/s,
Q->P = 1.5m/s,
\(Speed = \frac{Distance}{Time} \)
Inversely,
\(Time = \frac{Distance}{Speed}\)
Total time taken = 64s.
Therefore,
\(\frac{x}{2.5} + \frac{y}{1.5} + \frac{y}{2.5} + \frac{x}{1.5} = 64,\)
Rearranging,
\((x+y)(\frac{1}{2.5} + \frac{1}{1.5}) = 64\)
\((x+y) = 60\)
Bunuel
In an industry, the raw materials and the finished goods are transported on the conveyor belt. There are two conveyor belt, one for carrying parts from P to point Q and another for carrying parts from R to point Q. P, Q and R in that order are in a straight line. Sometimes, the belt serves to transport cart, which can themselves move with respect to the belts. The two belts move at a speed of 0.5 m/s and the cart move at a uniform speed of 2 m/s with respect to the belts. A cart goes from point P to R and back to P taking a total of 64 s. Find the distance PR in meters. Assume that the time taken by the cart to turn back at R is negligible?
A. 48
B. 54
C. 60
D. 64
E. 72
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