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A particle is thrown in upward direction with initial velocity $V_0$. It crosses point P at height h at time $t_1$ and $t_2$ so $t_1 t_2$ =
$ h = V_0 t - {1 \over 2 } gt^2 $ $ \therefore {1 \over 2 } gt^2 - V_0 t + h = 0 $ $ \therefore t^2 + { 2V_0 \over g } - { 2h \over g } = 0 $ There are two real values of t .
Slope of the velocity-time graph gives............ of a moving body.
The slope of a velocity-time graph represents the acceleration of the moving body. This is because acceleration is defined as the rate of change of velocity with respect to time. Therefore, the slope of the velocity-time graph is the acceleration. Hence, the correct option is o2: acceleration.
The intercept of the velocity-time graph on the velocity axis gives.
The intercept of the velocity-time graph on the velocity axis represents the velocity when time is zero. This is defined as the initial velocity ( ext{u}). Mathematically, it is the value of the velocity at the point where the time ( ext{t}) is zero.
In a uniformly accelerated motion the slope of velocity - time graph gives ....
In a uniformly accelerated motion, the slope of the velocity-time graph gives the acceleration ( ext{a}). This is because acceleration is defined as the rate of change of velocity with respect to time, which is represented by the slope of the velocity-time graph.
The area covered by the curve of V – t graph and time axis is equal to magnitude of ..
The area covered by the curve of the velocity-time (V-t) graph and the time axis represents the displacement ( ext{s}). This is because displacement is the integral of velocity with respect to time, which geometrically corresponds to the area under the velocity-time graph.
An object moves in a straight line. It starts from the rest and its acceleration is $2ms^{–2}$. After reaching a certain point it comes back to the original point.In this movement its acceleration is $ -3ms^{-2} $. Till it comes to rest. The total time taken for the movement is 5 second. Calculate the maximum velocity.
If maximum velocity is V $ V = a_1 t_1 \,and\, V =a_2 t_2 $ $ T = t _1 + t_2 = { v \over a_1 } + { v \over a_2 } $ $ V = { a_1 a_2 T \over a_1 + a_2 } $
Particles A and B are released from the same height at an interval of 2s. After some time t the distance between A and B is 100m. Calculate time t.
A particle is moving in a circle of radius R with constant speed. It covers an angle $ \theta $ in some time interval. Find displacement in this interval of time
$ \triangle S = \sqrt { R^2 + R^2 - 2R^2 cos \theta } $ $ \triangle S= \sqrt { 2R^2 - 2R^2 Cos \theta } = \sqrt { 2R^2 (1 - Cos \theta ) }$ $ = 2 R sin { \theta \over 2 } $
Angle of projection, maximum height and time to reach the maximum height of a particle are , H and tm respectively. Find the true relation.
To find the time to reach the maximum height, we use the kinematic equation for a projectile. The time to reach the maximum height (tm) is given by: \[ t_m = rac{u ext{sin} heta}{g} \\] The maximum height (H) is given by: \[ H = rac{u^2 ext{sin}^2 heta}{2g} \\] Rearranging to solve for tm, we get: \[ t_m = rac{u ext{sin} heta}{g} = rac{ ext{sin} heta}{g} imes rac{u^2 ext{sin} heta}{2u} \\] \[ t_m = rac{2H}{g} imes rac{1}{u} \\] Simplifying, we get: \[ t_m = rac{2H}{g} = rac{2H}{g} \\] Therefore, the correct option is o2: \[ t_m = rac{ ext{2H}}{g} \\]
A freely falling object travels distance H. Its velocity is V. Hence, in travelling further distance of 4H its velocity will become ....
$ y = V_0 t + { 1 \over 2 } gt^2 $ $ y = gt \times {t \over 2 } - {1 \over 2 } g \left( { t \over 2 } \right) ^2 $
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.