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A car is moving with a velocity (3.0 m/s)x + (1.0 m/s)y and 3.0 seconds later its velocity is (6.0m/s)x - (3.0 m/s)y. What is the direction of the average acceleration of the car?

User Vtrubnikov
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Final answer:

The average acceleration's direction is derived from the acceleration components and is located in the second quadrant, indicating a direction from the positive x-axis to the negative y-axis.

Step-by-step explanation:

The direction of the average acceleration of the car can be found using the change in velocity over the time interval. The initial velocity is given as vi = (3.0 m/s)i + (1.0 m/s)j, and the final velocity after 3.0 seconds is vf = (6.0 m/s)i - (3.0 m/s)j. To find the average acceleration, aavg, we calculate the change in velocity, Δv = vf - vi, and divide it by the time interval, Δt (which is 3.0 seconds).

The change in velocity is:

  • Δvx = 6.0 m/s - 3.0 m/s = 3.0 m/s
  • Δvy = -3.0 m/s - 1.0 m/s = -4.0 m/s

The average acceleration components are:

  • aavg,x = Δvx / Δt = 3.0 m/s / 3.0 s = 1.0 m/s2
  • aavg,y = Δvy / Δt = -4.0 m/s / 3.0 s = -1.33 m/s2

Therefore, the direction of the average acceleration is given by the vector sum of these components, which would be in the second quadrant of an x-y coordinate system, since the acceleration has a positive x-component and a negative y-component. The direction is given by the angle θ, which can be calculated using the arctan of the y-component over the x-component, θ = arctan(aavg,y/aavg,x). In this case, θ = arctan(-1.33/1.0) which indicates a direction that is clockwise from the negative y-axis towards the positive x-axis.

User Suresh Chaudhary
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