Practice free Motion in a Straight Line (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.
Motion of a particle is described by an equation $ \upsilon = (A + y ) ^{1 /2 } $
where v, y and A are velocity distance and a constant respectively. Find the
acceleratrion of the particle.
$ V = ( A + y ) ^{1 /2 } $ $ \therefore a = { dv \over dt } = { dv \over dy } . { dy \over dt } $
The minumum distance in which a car can be stopped is x. The velocity of the
car is V. If the velocity is 2V then find the stopping distance.
$ s = { v^2 \over 2 a } $
A goods train is moving with constant acceleration. when engine passes through a signal its speed is U. Midpoint of the train passes the signal with speed V. What will be the speed of the last wagon ?
For a train moving with constant acceleration, the speed of the last wagon can be found using the kinematic equations. Given that the speed of the engine when it passes the signal is U and the speed at the midpoint is V, we can use the relationship between these speeds to find the speed of the last wagon. The correct formula is $ ext{speed} = \\sqrt{2V^2 - U^2} $, which matches option o4.
The area under acceleration versus time graph for any time interval represents
The area under an acceleration vs. time graph represents the change in velocity over a given time interval. This is derived from the definition of acceleration, which is the rate of change of velocity. Therefore, integrating acceleration with respect to time gives the change in velocity. Hence, option o3 is correct.
x and y co-ordinates of a particle moving in x-y plane at some instant are $ x = 2 t^2 $ and $ y = {3 \over 2 } t^2 $ . Calculate y co-ordinate when its x coordinate is 8 cm
$ x = 2t^2 = 8 $ $ \therefore t = 2s $ $ y = { 3 \over 2 } t^2 = {3 \over2 } (2)^2 = 6m $
A particle in xy plane is governed by $ x = A cos \omega t, y = A (1 – sin \omega t)$ . A and $ \omega $ are constants. What is the speed of the particle.
$$ x = A coswt $$ $$ y = A ( i - sin wt ) $$ $$ V \alpha = { dx \over dv } = - A \omega sin wt $$ $$Vy = - A \omega cos wt $$ $$ V = \sqrt { Vx^2 + Vy^2 } $$
If a particle is moving along straight line with increasing speed, then
If the speed of body is increasing then acceleration is in the direction of velocity.It may be positive or negative. If acceleration is in negative direction then acceleration is increasing but in negative side, so it will be called as decreasing.
If magnitude of average speed and average velocity over a time interval are same, then
The magnitude of average speed and average velocity can only be equal if object moves in a straight line without turning back. In that condition distance will be equal to displacement.
The position of a particle moving along x-axis is given by x = 10t – 2t2. Then the time (t) at which it will momently come to rest is
x = 10t – 2t2
v = 0, at the time of coming to rest, so
10 – 4t = 0
t = 2.5 s
The position x of particle moving along x-axis varies with time t as
x = Asin (ωt)
where A and ω are positive constants. The acceleration a of particle varies with its position (x) as
The given equation represents a simple harmonic motion, where x = A sin(ωt) is the displacement. The acceleration (a) is the second derivative of displacement with respect to time, which is a = -ω^2 x. Hence, the correct option is (2) a = -ω^2 x.
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