Answer :

(The ‘average’ stands for average of the quantity over the time interval t_{1} to t_{2})

(a) False

If the velocity of particle is v (t_{1}) at time t = t_{1} and velocity is v (t_{2}) at time interval t = t_{2} the average velocity of particle is denoted by

v_{average} = (1/2) (v (t_{1}) + v (t_{2}))

Only in case of uniform motion that is when acceleration of the particle is constant or the rate of change of velocity is fixed, in case of arbitrary motion when acceleration may or may not be constant the above relation is incorrect

(b) True

We know the average velocity for any type of motion is given by v_{average} = total displacent / total time

Now r(t_{1}) and r(t_{2}) are the position vector of particle at time interval t = t_{1} and t = t_{2} respectively so r(t_{2}) - r(t_{1}) is the change in position of particle or the displacement of the particle and ( t_{2} – t_{1}) is the time taken for that displacement so [r(t_{2}) - r(t_{1}) ] /( t_{2} – t_{1}) denotes the average velocity of the particle v_{average}

(c) False

Here v (0) denotes the velocity of the particle at time interval, t = 0 i.e. it denotes initial velocity of the particle v(t) is the velocity of particle at any time interval t seconds and a is the acceleration, so it is the first equation of motion v = u + at

Where v is the final velocity after time t, u is the initial velocity and a is the acceleration, but this equation is valid only in case of uniform motion i.e. when acceleration of particle is constant but in case of arbitrary motion it may or may not be constant so above equation is not always valid.

(d) False

Here r (0) denotes the position of particle at time interval t = 0 , r (t) is the position of particle at time interval t , a is the acceleration of particle

r (t) = r (0) + v (0) t + (1/2) at^{2} can be rearranged as

r (t) - r (0) = v (0) t + (1/2) at^{2} so r (t) - r (0) is the change in position of particle or displacement v (0) is the velocity of particle at time interval t = 0 or initial velocity of the particle i.e. this equation represents the equation of motion

S = ut + ( �)at^{2} where S is the displacement of particle u is the initial velocity of the particle a is acceleration and t is time but the above equation is valid only in case of uniform motion i.e. when acceleration of particle is constant but in case of arbitrary motion it may or may not be constant so above equation is not always valid.

(e) True

We know the average acceleration for any type of motion is given by a_{average} = change in velocity / total time

Now v(t_{1}) and v(t_{2}) are the velocity of particle at time interval

t = t_{1} and t = t_{2} respectively so v(t_{2}) - v(t_{1}) is the change in velocity of particle and ( t_{2} – t_{1}) is the time taken for velocity to change so [v(t_{2}) - v(t_{1}) ] /( t_{2} – t_{1}) denotes the average acceleration of the particle a_{average}

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