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A robot used in a pharmacy picks up a medicine bottle at t=0. It accelerates at 0.15 m/s2 for 7.0 s , then travels without acceleration for 64 s and finally decelerates at -0.50 m/s2 for 2.1 s to reach the counter where the pharmacist will take the medicine from the robot. From how far away did the robot fetch the medicine?

Answer :

Answer:

Explanation:

distance while accelerating

d = ½at² = ½(0.15)7.0² = 3.675 m

velocity at the end of 7 s

v = at = 0.15(7.0) = 1.05 m/s

distance at constant velocity

d = vt = 1.05(64) = 67.2 m

distance while decelerating

d = ut + ½at² = 1.05(2.1) + ½(-0.50)2.1² = 1.1025 m

velocity at the end of deceleration

v = u + at = 1.05 + (-0.5)2.1 = 0 m/s

total distance traveled

d = 3.675 + 67.2 + 1.1025 = 71.9775 m

rounding to the two significant figures of the question numerals

d = 72 m

that's a HUGE pharmacy.

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