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A particle is projected from the ground, where equation of trajectory is given by \mathrm{y(t)=a t+b t^2} where a and b are constant. Then horizontal rate of velocity is-

Option: 1

Independent of time


Option: 2

Dependent of time


Option: 3

bt


Option: 4

a+bt


Answers (1)

best_answer

Trajectory is given by \mathrm{y(t)=a t+b t^2}

Horizontal rate is given by \mathrm{\frac{d y(t)}{d t}} is equal to 

\mathrm{\frac{d y(t)}{d t}=a+2 b t} ------------(1)

Now, the Horizontal rate of velocity 

\mathrm{\frac{d^2 y(t)}{d t^2}=0+2 b}                                            

    \mathrm{\frac{d^2 y(t)}{d t^2}=2 b}  Independent of time

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