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An object moves with constant acceleration $ a $ . Which of the following expressions are also constant?
A) $ \dfrac{{d\left| v \right|}}{{dt}} $
B) $ \left| {\dfrac{{dv}}{{dt}}} \right| $
C) $ \dfrac{{d\left( {{v^2}} \right)}}{{dt}} $
D) $ \dfrac{{d\left( {\dfrac{v}{{\left| v \right|}}} \right)}}{{dt}} $

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Last updated date: 13th Jun 2024
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Answer
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Hint: Acceleration of an object is defined as the rate of change of velocity of an object with respect to time. Velocity and acceleration of an object are vector quantities. If the acceleration of an object is constant it must move with constant velocity with respect to time.

Complete step by step solution:
We’ve been given that an object moves with constant acceleration $ a $ and we’ve been asked to find which expression from the option remains constant.
Since the acceleration of the object is constant, we can say that $ \vec a = {\text{constant}} $ . For the acceleration to be constant, both its magnitude and directions have to be constant. So we can write that the magnitude of the acceleration of the object $ \left| a \right| = {\text{constant}} $ .
Since $ \left| a \right| = \left| {\dfrac{{dv}}{{dt}}} \right| $ , and the magnitude of the acceleration i.e. the term on the left side is constant, the term on the right side of the equation will also be constant with time.
So option (B) is the correct choice.

Note:
Option (A) is incorrect because while the magnitude of the velocity is constant, its direction might change so the time derivative might have different directions as is the case in a circular motion. Option (C) is incorrect because there is no relation of acceleration with the time derivative of the square of the velocity which is given is mentioned in the option. Option (D) is also incorrect as it talks about the time derivative of the unit vector of velocity as $ \left( {\dfrac{v}{{\left| v \right|}}} \right) = \hat v $ which is related to the unit vector of acceleration and not acceleration itself.