A bus is moving on a straight road towards north with a uniform speed of 50 km/hour then it turns left through 90$^{\circ}$. If the speed remains unchanged after turning, the increase in the velocity of bus in the turning process is

Question

A ball is thrown vertically upwards with an initial velocity of $ 20 \, \text{m/s} $. How high will the ball rise? (Take $ g = 10 \, \text{m/s}^2 $)

Answer 1

To determine the increase in velocity of the bus as it turns, we need to consider the initial and final velocities as vectors.

The initial velocity of the bus is directed towards north at a speed of 50 km/h. After turning 90^{\circ} to the left, the velocity is directed towards west, with the same speed of 50 km/h.

We can represent the initial and final velocities as vectors:

Initial velocity, \vec{v}_{\text{initial}} = 50 \, \text{km/h} \, \hat{j} (north direction)
Final velocity, \vec{v}_{\text{final}} = 50 \, \text{km/h} \, \hat{i} (west direction)

We can find the change in velocity by vector subtraction:

The change in velocity \Delta \vec{v} = \vec{v}_{\text{final}} - \vec{v}_{\text{initial}}

Substituting the values, we get:

\Delta \vec{v} = 50 \, \hat{i} \, - \, 50 \, \hat{j}

The magnitude of this change in velocity is given by:

|\Delta \vec{v}| = \sqrt{50^2 + (-50)^2} = \sqrt{2500 + 2500} = \sqrt{5000} = \sqrt{100 \times 50} = 10\sqrt{50} = 70.7 \, \text{km/h}

Since the change in velocity forms an angle of 45^{\circ} with both the initial and final velocities, it is directed towards the south-west direction.

Thus, the increase in the velocity of the bus during the turning process is 70.7 km/hr along south-west direction.

Answer 2