Question

An aeroplane is flying with a uniform speed of 150 km/h along the circumference of a circle. The change in its velocity in half the revolution (in km/h) is:
A.) 150
B.) 100
C.) 200
D.) 300
E.) 50

Answer 1

Hint: The aeroplane is moving in a circular path, therefore, the velocity at any point in a circular path is taken at the tangent of the circle. If velocity is taken as ${v_1}$ at one point on the circle, then at the halfway of the circle, take the velocity as ${v_2}$. Magnitude of both the velocities is the same but the direction is different as shown in the figure.

Complete step-by-step answer:
Since the aeroplane is moving at the uniform speed in circular motion, ${v_1}$ is the same as ${v_2}$ , but their direction is opposite as the velocity is taken at the tangents.

We need to calculate the velocity difference in half the revolution, which will be the difference of both the velocities. Since velocity is a vector quantity, its direction also needs to be specified and therefore, negative signs would be taken into consideration. Let the velocity difference be
$\Delta v$
Therefore, \[\Delta v{\text{ }} = {\text{ }}{v_1}-{v_2}\]
Here, $v_1$ is given as 150 km/h
The value of ${v_2}$ will be -150 km/h.

Now change in velocities is given by \[\Delta v{\text{ }} = {\text{ }}{v_1}-{v_2}\]
Substituting the values, we get

 $ \Rightarrow $ \[\Delta v = 150 - ( - 150)\]
$ \Rightarrow $ $\Delta v = 150 + 150$
$ \Rightarrow \Delta v = 300$

Therefore, the change in velocity in half the revolution is 300 km/h.

Hence, option (D) is the correct answer.

Note: The confusing thing here was only that after half the revolution, the sign change should be considered, otherwise the answer could come zero. Also this should be known that the velocities are taken along the tangent in a circular path, in that way the velocity after half the revolution becomes opposite in direction but has the same magnitude if travelling in uniform speed.