A body is moving according to the equation $x = at + b{t^2} - c{t^3}$ where $x = $ displacement and $a,\;b$ and $c$ are constants. The acceleration of the body is
- A$a + 2bt$
- B$2b + 6ct$
- C$2b - 6ct$
- D$3b - 6c{t^2}$
Similar Questions
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A particle executes the motion described by $x(t) = x_0 (1 - e^{-\gamma t} )$ ; જ્યાં $t\, \geqslant \,0\,,\,{x_0}\, > \,0$.
$(a)$ Where does the particle start and with what velocity ?
$(b)$ Find maximum and minimum values of $x(t),\, v(t)$ $a(t)$. Show that $x(t)$ and $a(t)$ increase with time and $v(t)$ decreases with time.
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$(a)$ Where does the particle start and with what velocity ?
$(b)$ Find maximum and minimum values of $x(t),\, v(t)$ $a(t)$. Show that $x(t)$ and $a(t)$ increase with time and $v(t)$ decreases with time.
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A body starts from the origin and moves along the $X-$axis such that the velocity at any instant is given by $(4{t^3} - 2t)$, where $t$ is in sec and velocity in$m/s$. What is the acceleration of the particle, when it is $2\, m$ from the origin..........$m/{s^2}$
A body starts from the origin and moves along the $X-$axis such that the velocity at any instant is given by $(4{t^3} - 2t)$, where $t$ is in sec and velocity in$m/s$. What is the acceleration of the particle, when it is $2\, m$ from the origin..........$m/{s^2}$
A particle moves a distance $x$ in time $t$ according to equation $x = (t + 5)^{-1}$ The acceleration of particle is proportional to
A particle moves a distance $x$ in time $t$ according to equation $x = (t + 5)^{-1}$ The acceleration of particle is proportional to
- [AIPMT 2010]