Practice free Motion in a Lane (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
$ \vec A and \vec B $ are two vectors $ \hat U_A = \hat U_B $ Now find the true option
$ \vec A + \vec B + \vec C = \hat j $
Angle of projection of a projectile with horizonal line is $ \theta $ at time t = 0, After what time the angle will be again $ \theta $ ?
The time of flight for a projectile is given by \( \frac{2V_0 \sin \theta}{g} \). Since the angle of projection \( \theta \) will repeat at half the time of flight, the time when the angle will be \( \theta \) again is \( \frac{2V_0 \sin \theta}{g} \).
A particle is projected with initial speed of $V_0$ and angle of $ \theta $ . Find the horizontal displacement when its velocity is perpendicular to initial velocity.
$ \vec V_0 = V_0cosθ \hat i + V_0 sinθ \hat j$ $ \vec V = V_0 cosθ \hat i + (V_0sm \theta – gt) \hat j$ $ \vec V_0 . \vec V = 0 $ $ \therefore t = { V_0 \over gsin \theta } $ Now find x
Intial angle of a projectile is $\theta $ and its initial velocity is $V_0$. Find the angle of velocity with horizontal line at time t.
At time t $ V_ x = V_0 = V_0 cos \theta $
$ V_y = Visinθ – gt$
$ tan \alpha = { Vy \over Vx } = {V_0sin \theta – gt \over V_0cos \theta } $
A stone is projected with an angle $ \theta $ and velocity $V_0$ from point P. It strikes the ground at point Q. If the both P and Q are on same horizontal line, then find average velocity.
For a projectile motion where the initial and final points are on the same horizontal line, the average velocity is the horizontal component of the initial velocity, which is \( V_0 \cos \theta \).
Angle of projection of a projectile is changed, keeping initial velocity constant. Find the rate of change of maximum height. Range of the projectile is R.
A body travelling in a circle at constant speed.
When a body travels in a circle at constant speed, it is undergoing circular motion. In circular motion, even though the speed is constant, the direction of the velocity is constantly changing. This change in direction means there is an acceleration, called centripetal acceleration, which acts towards the center of the circle. Therefore, the body has an inward radial acceleration.
If f is the frequency of a body moving in a circular path with constant speed. a is its centrifugal acceleration, so.
Centrifugal acceleration $ a_r = { v^2 \over r } = \left( { 2 \pi r \over T } \right) ^2 { 1 \over r } = 4 \pi^2 r f^2 $
Following question is Assertion - Reason type question. Choose Assertion : At the highest point of projectile motion the velocity is not zero. Reason : Only the verticle component of velocity is zero. Where as horizontal component still exists
In projectile motion, at the highest point, the vertical component of the velocity is zero but the horizontal component of the velocity still exists. Thus, the total velocity is not zero. Hence, both the assertion and reason are true, and the reason correctly explains the assertion.
A particle is moving in a circle of radius R with constant speed. The time period of the particle is T Now after time t = T/6 .Average velocity of the particle is
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.