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4..What is the cyclist's average velocity if his position changes from 30 m, North to 50 m, North during a 5 second time interval?
7. As she climbs a hill, a cyclist slows down from 25 mi/hr to 6 mi/hr in 10 seconds. What is her deceleration?
4..What is the cyclist's average velocity if his position changes from 30 m, North to 50 m, North during a 5 second time interval?
7. As she climbs a hill, a cyclist slows down from 25 mi/hr to 6 mi/hr in 10 seconds. What is her deceleration?
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Answer:
4.
Given:
Initial position: 30 m, N
Final position: 50 m, N
Time: 5 s
Formula:
velocity= displacement
time
displacement= 50 m - 30 m
= 20 m
Solution:
velocity= 20 m
5 m
velocity= 4 m/s, North
The cyclist's average velocity is 4 m/s, North
7.
Given:
Initial velocity: 25 mi/hr
Final velocity: 6 mi/hr
Time: 10 s
Formula:
deceleration= change in velocity
time
Solving:
change in velocity= 6 mi/hr - 25 mi/hr
= -19 mi/hr
deceleration= -19 mi/hr
10 s
deceleration= - 1.9 mi/hr/s
The cyclist's average deceleration is -1.9 mi/hr/s
Explanation:
i hope it helps!
I'm not sure if correct yung sa deceleration.. pa correct nalang kung mali :)
pa brainliest po