Answered

Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Ask your questions and receive detailed answers from professionals with extensive experience in various fields. Our platform offers a seamless experience for finding reliable answers from a network of knowledgeable professionals.

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?

Sagot :

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.

We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.