For this problem, we use the derived equations for rectilinear motion at constant acceleration. The equations used for this problem are:
a = (v - v₀)/t
2ax = v² - v₀²
where
a is the acceleration
x is the distance
v is the final velocity
v₀ is the initial velocity
t is the time
The solution is as follows;
a = (60mph - 30 mph)/(3 s * 1 h/3600 s)
a = 36,000 mph²
2(36,000 mph²)(x) = 60² - 30²
Solving for x,
x = 0.0375 miles
Answer:
Explanation:
We shall express each displacement vectorially , i for each unit displacement towards east , j for northward displacement and k for vertical displacement .
14 m due west = - 14 i
22.0 m upward in the elevator = 22 k
12 m north = 12 j
6.00 m east = 6 i
Total displacement = - 14 i + 22 k + 12 j + 6 i
D = - 8 i + 12 j + 22 k
magnitude = √ ( 8² + 12² + 22² )
= √ ( 64 + 144 + 484 )
= √ 692
= 26.3 m
Net displacement from starting point = 26.3 m .
<span>Assuming the car is travelling in the same direction for the entire hour, the acceleration is zero.</span>
Explanation:
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