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JEE PHYSICS MCQ: MOTION IN PLANE

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From a point on the ground, a particle is projected slantwise with some initial velocity making an acute angle with the horizontal. During its flight, the dot product of its velocity and acceleration vectors
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Correct option: A
Continuously increases
Option D (First decreases and then increases) is correct because the dot product of velocity and acceleration changes sign as the angle between them changes from obtuse to acute.

More Questions from MOTION IN PLANE

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Given, P⃗=A⃗+B⃗  and P=A+B.\mathrm{\vec P} = \mathrm{\vec A} + \mathrm{\vec B}\,\,\mathrm{and}\,P = A + B. The angle between A⃗ and B⃗\mathrm{\vec A}\,and\,\mathrm{\vec B} is Assertion: The necessary condition for motion in two dimensions, velocity and acceleration vectors should have any angle between 0∘0^{\circ} and 90∘90^{\circ} between them.
Reason: For two-dimensional or three-dimensional motion, velocity and acceleration of an object works in opposite direction.
If (a+3b)⋅(7a−5b)=0\mathrm{If}\,(\mathbf{a}+3 \mathbf{b}) \cdot(7 \mathbf{a}-5 \mathbf{b})=0 and
(a−4b)⋅(7a−2b)=0(\mathbf{a}-4 \mathbf{b}) \cdot(7 \mathbf{a}-2 \mathbf{b})=0. Then, the angle between a\mathbf{a} and b\mathbf{b} is
At the top of the trajectory of a projectile, the direction of its velocity and acceleration are A ball is thrown at an angle θ\theta and another ball is thrown at an angle (90∘−θ)\left(90^{\circ}-\theta\right) with the horizontal direction from the same point with velocity 39.2 ms−139.2 \mathrm{~ms}^{-1}. The second ball reaches 50 m50 \mathrm{~m} higher than the first ball. Find their individual heights. Take g=9.8 ms−2{g}=9.8 \mathrm{~ms}^{-2}. Assertion: The drover in a vehicle moving with a constant speed on a straight road is in a non-inertial frame of reference.

Reason: A reference frame in which Newton's laws of motion are applicable is non-inertial.
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