Q30E - University Physics with Modern Physics

Question

A railroad flatcar is traveling to the right at a speed of 13.0m/s relative to an observer standing on the ground. Someone is riding a motor scooter on the flatcar (Fig. E3.30). What is the velocity (magnitude and direction) of the scooter relative to the flatcar if the scooter’s velocity relative to the observer on the ground is (a) 18m/s to the right? (b) 3.0m/s to the left? (c) zero?

Step-by-Step Solution

Answer

1. The velocity of scooter relative to flatcar is 5m/s when the velocity of the scooter relative to the observer is 18m/s.
2. The velocity of the scooter relative to the flatcar is -16m/s when the velocity of the scooter relative to the observer is 3m/s to the left.
3. The velocity of the scooter relative to the flatcar is -16m/s when the velocity of the scooter relative to the observer is zero.

Step 1: Identification of given data:

The given data can be listed below.

• The velocity of the scooter relative to ground is v_sg = 13m/s.

Step 2: Concept/Significance of relative velocity:

Relative velocity is defined as the velocity of an object relative to another observer. It is the time rate of change of relative position of one object with respect to another object.

 

A frame of reference is a reference against which motion is examined. The speed of a body from a point of reference is its relative velocity.

Step 3: (a) Determination of the velocity (magnitude and direction) of the scooter relative to the flatcar if the scooter’s velocity relative to the observer on the ground is 18m/s to the right:

The velocity of the scooter relative to the flatcar is given by,

$$\begin{gathered} v_{so}=v_{sf}+v_{sg} \\ {v}_{sf}=v_{so}-v_{sg} \end{gathered}$$

Here, v_so is the velocity of the scooter relative to the observer, and v_sg is velocity of the scooter relative to the ground.

Substitute all the values in the above,

$\begin{array}{rcl}{v}_{sf}& =& 18m/s-13m/s\\ & =& 5m/s\end{array}$

Thus, the velocity of scooter relative to flatcar is 5 m/s when velocity of scooter relative to observer is 18m/s.

Step 4: (b) Determination of the velocity (magnitude and direction) of the scooter relative to the flatcar if the scooter’s velocity relative to the observer on the ground is 3 m/s to the right:

The velocity of the scooter relative to the flatcar is given by,

$$\begin{gathered} v_{so}=v_{sf}+v_{sg} \\ {v}_{sf}=v_{so}-v_{sg} \end{gathered}$$

Here, v_so is the velocity of the scooter relative to the observer and v_sg is the velocity of the scooter relative to the ground.

Substitute all the values in the above.

$\begin{array}{rcl}{v}_{sf}& =& -3m/s-13m/s\\ & =& -16m/s\end{array}$

Thus, the velocity of the scooter relative to the flatcar is -16m/s when the velocity of the scooter relative to the observer is 3m/s to the left.

Step 5: Determination of the velocity (magnitude and direction) of the scooter relative to the flatcar if the scooter’s velocity relative to the observer on the ground is zero:

The velocity of scooter relative to the flatcar is given by,

$$\begin{gathered} v_{so}=v_{sf}+v_{sg} \\ {v}_{sf}=v_{so}-v_{sg} \end{gathered}$$

Here, v_so is the velocity of scooter relative to the observer, and v_sg is velocity of scooter relative to the ground.

Substitute all the values in the above,

$\begin{array}{rcl}{v}_{s/f}& =& -0-13m/s\\ & =& -13m/s\end{array}$

Hence, the velocity of the scooter relative to the flatcar is -16m/s when the velocity of the scooter relative to observer is zero.