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 oppo

Question

2 years ago

11

site direction. What was the average acceleration of the particle during this 9.8 s interval

1 answer:

dem82 [27] · Answer 1 · 2023-06-12T19:05:26+03:00

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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