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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 $
A ball is thrown vertically upward direction. Neglecting the air resistance velocity of the ball in air will
When a ball is thrown vertically upwards, its velocity decreases as it ascends due to the force of gravity acting in the opposite direction. At the peak of its ascent, the velocity becomes zero. As the ball starts descending, gravity causes its velocity to increase in the downward direction. Therefore, the velocity of the ball decreases when it is going up. Hence, the correct option is o2.
Two particles P and Q get 5 m closer each second while travelling in opposite direction. They get 1 m closer each second while travelling in same direction. The speeds of P and Q are respectively ...
$$ V_1 + V_2 = 5 $$ $$ V_1 -V_2 = 1 So 2V_1 = 6 $$ $$ V_1 = 3 m/s $$
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