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Answer :
The runner's acceleration is -2 m/s², indicating deceleration at a rate of 2 meters per second squared over a period of 2 seconds.
The acceleration of a runner who goes from 6 m/s to 2 m/s in 2 seconds can be calculated using the formula for acceleration which is Δv/Δt, where Δv is the change in velocity and Δt is the change in time. In this case, Δv = final velocity - initial velocity = 2 m/s - 6 m/s = -4 m/s (the negative sign indicates deceleration). The time interval Δt is 2 seconds. Thus, acceleration a is calculated as:
a = Δv/Δt = (-4 m/s)/2 s = -2 m/s².
This indicates that the runner is decelerating at a rate of 2 meters per second squared.
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Rewritten by : Barada
Acceleration = Δv/Δt where v is velocity and t is time
So the acceleration is (6-2)/2= 2m/s^2
So the acceleration is (6-2)/2= 2m/s^2
