The average acceleration of the particle in the time interval of 9.8 s is 10.204 m/s² opposite to the direction of motion.
Given:
Speed of particle, v₁ = 80 m/s (in positive x-direction)
Speed of particle, v₂ = -20 m/s (in opposite direction)
Time interval, Δt = 9.8 s
Calculation:
We know that, the average acceleration is given as:
a_avg = (v₂ - v₁)/ Δt - ( 1 )
Applying values in above equation we get:
a_avg = (v₂ - v₁)/ Δt
= (-20 m/s - 80 m/s) / (9.8 s)
= -10.204 m/s²
Therefore the average acceleration of the particle in the time interval Δt is 10.204 m/s² opposite to the direction of motion.
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The depth considered is
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Converting to atmosphere
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Answer:
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