On the Acme W&W company production line, parts have to be transported 100 m by c
ID: 1407611 • Letter: O
Question
On the Acme W&W company production line, parts have to be transported 100 m by conveyor belt to a machine area. The parts move at a constant speed of 2.5 m/s. How much time does it take to move a part 100m?
On the same conveyor belt, the part must slow down to 0.2 m/s over a distance of 0.9m once it reaches the machine area. The original speed was 2.5 m/s. What is the average acceleration while slowing down? How much time does it take to slow down?
When leaving the machine area described earlier, the part reverses direction and starts with a velocity of 0.2 m/s. it is accelerated at 1.5m/s^2 for its journey back a distance of 5m. How long does it take the part to travel the 5m distance? What is the final velocity of the part?
Explanation / Answer
initial speed = vo = 2.5 m/s
acceleration a = 0
time = D/vo = 100/2.5 = 40 s
----------------
for the last 0.9 m
iitial speed = vo = 2.5 m/s
final speed = vf = 0.2 m/s
from equations of motion
vf^2-vo^2 = 2*a*x
0.2^2 - 2.5^2 = 2*a*0.9
a = -3.45 m/s^2 <<<-------answer
time t = (vf-vo)/a
t = (0.2-2.5)/3.45
t = 0.67 s <<<-------answer
++++++++++++
for reverse part
initial speed = vo = 0.2 m/s
distance travelled x = 5 m
acceleration = a = 1.5 m/s^2
x = vo*t + 0.5*g*t^2
5 = (0.2*t) + (0.5*1.5*t^2)
t = 2.45 s <<<----answer
vf^2 - vo^2 = 2*a*x
vf^2 - 0.2^2 = 2*1.5*5
vf = 3.87 m/s <<<<----answer
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.