Question

A girl at an airport rolls a ball north on a moving walking that moves east. If the ball’s speed with respect to the walkway is 0.15m/s and the walkway moves at a speed of 1.50m/s, what is the velocity of the ball relative to the ground?

Answer 1

We have to add two vectors.

Vector #1: 0.15 m/s north

Vector #2: 1.50 m/s east

Their sum:

Magnitude: √(0.15² + 1.50²)

Magnitude = √(0.0225+2.25)

Magnitude = √2.2725

Magnitude = 1.5075 m/s

Direction = arctan(0.15/1.50) north of east

Direction = 5.71° north of east

Answer 2

The velocity of the ball relative to the ground is 1.5m/s at 5.7° North East.

Given the data in the question;

• Velocity of walkway with respect to the ground along east direction; $v_{wg} = 1.50m/s$

• Velocity of the ball with respect to the walkway along north direction; $v_{bw}=0.15m/s$

• Velocity of the ball with respect to the ground; $v_{bg} = \ ?$

To determine the velocity of the ball relative to the ground.

From Parallelogram law:  

Resultant velocity; $v^2 = v_1^2 + v_2^2$

So, $v = \sqrt{v_1^2 + v_2^2}$

In this case, $v_{bg} = \sqrt{v_{wg}^2 + v_{bw}^2}$

We substitute in our values

$v_{bg} = \sqrt{(1.5m/s)^2 + (0.15m/s)^2}\\\\v_{bg} = \sqrt{2.25m^2/s^2 + 0.0225m^2/s^}$

$v_{bg} = \sqrt{2.2725m^2/s^2$

$v_{bg} = 1.5m/s$

For the direction

$tan\theta = \frac{v_{bw}}{v_{wg}} = \frac{0.15m/s}{1.5m/s} = 0.1$

$\theta = tan^{-1}(0.1)\\\\\theta = 5.7^o$

Therefore, the velocity of the ball relative to the ground is 1.5m/s at 5.7° North East.

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