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Q3.25(a):

On a long horizontally moving belt (Fig. 3.26), a child runs to and fro with a speed $9 km h^{-1}$ (with respect to the belt) between his father and mother located $50 m$ apart on the moving belt. The belt moves with a speed of $4 km h^{-1}$. For an observer on a stationary platform outside, what is the (a) speed of the child running in the direction of motion of the belt ?.

Solution :

Step 1: Establishing Coordinate Systems and Variables

To determine the velocity of the child with respect to a stationary observer, we apply the principles of relative velocity in one dimension [Per the Galilean Transformation of velocities].

  • Velocity of the belt relative to the stationary ground ($v_{b}$): $+4 \text{ km h}^{-1}$
  • Velocity of the child relative to the belt ($v_{c/b}$): $+9 \text{ km h}^{-1}$ (running in the direction of the belt's motion)
  • Velocity of the child relative to the stationary ground ($v_{c}$): To be determined.

Step 2: Vector Addition of Relative Velocities

The velocity of an object $c$ with respect to a stationary frame $g$ is the vector sum of its velocity with respect to a moving frame $b$ and the velocity of that moving frame with respect to the stationary frame $g$. [By the Law of Addition of Velocities]:

$v_{c} = v_{c/b} + v_{b}$

Step 3: Algebraic Calculation

Substitute the given values into the equation:

$v_{c} = 9 \text{ km h}^{-1} + 4 \text{ km h}^{-1}$

$v_{c} = 13 \text{ km h}^{-1}$

Visual Representation of Vector Addition

Belt $v_{b}=4$ Child wrt Belt $v_{c/b}=9$ Resultant $v_{c}=13 \text{ km h}^{-1}$

Step 4: Verification and Unit Consistency

The calculation assumes the motion is strictly linear and collinear. Since both the belt and the child move in the positive direction relative to the ground, the magnitudes are added arithmetically.

Component Value
$v_{c/b}$ $9 \text{ km h}^{-1}$
$v_{b}$ $4 \text{ km h}^{-1}$
$v_{c}$ (Resultant) $13 \text{ km h}^{-1}$

Final Solution: The speed of the child running in the direction of motion of the belt, as observed from the stationary platform, is $13 \text{ km h}^{-1}$.


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