A 40.0 -kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 13
1 answer:
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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>