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Answer:
The time interval of acceleration for the bus is 2.20 seconds
Explanation:
Acceleration is the rate of change of velocity
→ 
where a is the acceleration, v is the final velocity, u is the initial velocity
and t is the time
The given is:
The uniform acceleration = -4.1 m/s²
The bus slows from 9 m/s to 0 m/s
We need to find the time interval of acceleration for the bus
Lets use the rule above
→ a = -4.1 m/s² , v = 0 m/s , u = 9 m/s
→ 
Multiply both sides by t
→ -4.1 t = -9
Divide both sides by -4.1
∴ t = 2.20 seconds
<em>The time interval of acceleration for the bus is 2.20 seconds</em>
Answer:
a=1.25m/s²
Explanation:
GIVEN DATA
vi=10m/s
vf=15m/s
S=5m
TO FIND
a=?
SOLUTION
by using third equation of motion
2as=(vf)²-(vi)²
2a(5m)=(15m/s)²-(10m/s)²
10m×a=225m²/s²-100m²/s²
10m×a=125m²/s²
a=
a=12.5m/s²