1.
You see an ant walking across a running treadmill. The ant is moving with a velocity of 0.1 m/s in the +x
direction. The treadmill is moving 0.5 m/s in the +y direction. What is the ant's resultant velocity from
where you are standing?

Since the ant is moving with a velocity of 0.1 m/s in the +x direction relative to the treadmill, and the treadmill is moving 0.5 m/s in the +y direction, the magnitude of the resultant velocity of the ant relative to the ground (since both directions are perpendicular) is thus

V = √(x² + y²)

= √[(0.1 m/s)² + (0.5 m/s)²]

= √[0.01 m²/s² + 0.25 m²/s²]

= √(0.26 m²/s²)

= 0.51 m/s.

The direction is Ф = tan⁻¹(y/x)

= tan⁻¹(0.5 m/s ÷ 0.1 m/s)

= tan⁻¹(5)

= 78.69°

≅ 78.7°

The ant's resultant velocity is 0.51 m/s at 78.7°

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1. You see an ant walking across a running treadmill. The ant is moving with a velocity of 0.1 m/s in the +x direction. The treadmill is moving 0.5 m/s in the +y direction. What is the ant's resultant velocity from where you are standing?

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1.
You see an ant walking across a running treadmill. The ant is moving with a velocity of 0.1 m/s in the +x
direction. The treadmill is moving 0.5 m/s in the +y direction. What is the ant's resultant velocity from
where you are standing?