Question

At a certain time a particle had a speed of 80 m/s in the positive x direction, and 9.8 s later its speed was 20 m/s in the opposite direction. What was the average acceleration of the particle during this 9.8 s interval

Answer 1

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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