Question 18: Motion in 1D

Question

Train A is moving along two parallel rail tracks towards north with speed $$72 \mathrm{~km} / \mathrm{h}$$ and train B is moving towards south with speed $$108 \mathrm{~km} / \mathrm{h}$$. Velocity of train B with respect to A and velocity of ground with respect to B are (in $$\mathrm{ms}^{-1}$$):

Accepted Answer

To find the velocity of Train B with respect to Train A, we have to subtract the velocity of Train A from the velocity of Train B, keeping in mind that they are moving in opposite directions. Since they are moving in opposite directions, the relative velocity is calculated by adding their magnitudes when converting into the same unit, which in this case is meters per second.
First, let's convert the speeds from km/h to m/s by multiplying by the conversion factor $$ \frac{1000 	ext{ m/km}}{3600 	ext{ s/h}} = \frac{5}{18} 	ext{ m/s} $$.
For Train A:
$$
v_A = 72 \frac{	ext{km}}{	ext{h}} 	imes \frac{5}{18} \frac{	ext{m/s}}{(	ext{km/h})} = 20 \frac{	ext{m}}{	ext{s}}
$$
For Train B:
$$
v_B = 108 \frac{	ext{km}}{	ext{h}} 	imes \frac{5}{18} \frac{	ext{m/s}}{(	ext{km/h})} = 30 \frac{	ext{m}}{	ext{s}}
$$
To find the velocity of B relative to A ($ v_{B/A} $), we consider the direction: Train B is moving towards the south and Train A is moving towards the north. Therefore, relative to Train A, Train B is moving even faster towards the south, we calculate:
$$
v_{B/A} = v_B + v_A = 30 \frac{	ext{m}}{	ext{s}} + 20 \frac{	ext{m}}{	ext{s}} = 50 \frac{	ext{m}}{	ext{s}}
$$
Since Train B is moving towards the south and Train A towards the north, we take the southward direction as negative in our coordinate system for this calculation. That means the velocity of B with respect to A is:
$$
v_{B/A} = -50 \frac{	ext{m}}{	ext{s}}
$$
Next, we calculate the velocity of the ground with respect to Train B ($ v_{ground/B} $). The ground is stationary, thus it has a velocity of 0 m/s in any direction. The velocity of an object with respect to another object moving is just the opposite of the second object's velocity. Thus:
$$
v_{ground/B} = -v_B = -30 \frac{	ext{m}}{	ext{s}}
$$
However, because we are considering the southward direction as negative, the negative of a southward velocity is a northward velocity. Hence we get:
$$
v_{ground/B} = 30 \frac{	ext{m}}{	ext{s}}
$$
So the velocity of Train B with respect to Train A is -50 m/s, and the velocity of the ground with respect to Train B is 30 m/s. Therefore, the correct answer is:
Option A: -50 m/s and -30 m/s. (Incorrect, because the velocity of ground with respect to B is positive in our chosen coordinate system)
Option B: -50 m/s and 30 m/s. (Correct)
Option C: -30 m/s and 50 m/s. (Incorrect)
Option D: 50 m/s and -30 m/s. (Incorrect)
Thus, the correct answer is Option B: -50 m/s and 30 m/s.

Answer

-50 and -30

Answer

-50 and 30

Answer

-30 and 50

Answer

50 and -30

Motion in 1D

Train A is moving along two parallel rail tracks towards north with speed $72 \mathrm{~km} / \mathrm{h} and train B is moving towards south with speed 108 \mathrm{~km} / \mathrm{h}. Velocity of train B with respect to A and velocity of ground with respect to B are (in \mathrm{ms}^{-1}$):

Choose the correct answer: