JEE MAIN - Physics (2019 - 9th April Evening Slot - No. 24)
The position of a particle as a function of time
t, is given by
x(t) = at + bt2 – ct3
where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be :
x(t) = at + bt2 – ct3
where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be :
$$a + {{{b^2}} \over {c}}$$
$$a + {{{b^2}} \over {4c}}$$
$$a + {{{b^2}} \over {3c}}$$
$$a + {{{b^2}} \over {2c}}$$
Explanation
x = at + bt2 – ct3
$$V = {{dx} \over {dt}} = a + 2bt - 3c{t^2}$$
$$a = {{dv} \over {dt}} = 2b - 6ct$$
Put acceleration = 0
$$ \Rightarrow t = {b \over {3c}}$$
Now V at t = $${b \over {3c}}$$
$$V = a + {{{b^2}} \over {3c}}$$
$$V = {{dx} \over {dt}} = a + 2bt - 3c{t^2}$$
$$a = {{dv} \over {dt}} = 2b - 6ct$$
Put acceleration = 0
$$ \Rightarrow t = {b \over {3c}}$$
Now V at t = $${b \over {3c}}$$
$$V = a + {{{b^2}} \over {3c}}$$
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