Question 71: Motion in 1D

Question

The position of a particle as a function of time
t, is given byx(t) = at + bt2 – ct3
where a, b and c are constants. When the
particle attains zero acceleration, then its
velocity will be :

Accepted Answer

x = at + bt
2
Ã¢ÂÂ ct
3
$$V = {{dx} \over {dt}} = a + 2bt - 3c{t^2}$$
$$a = {{dv} \over {dt}} = 2b - 6ct$$
Put acceleration = 0
$$ \Rightarrow t = {b \over {3c}}$$
Now V at t = $${b \over {3c}}$$
$$V = a + {{{b^2}} \over {3c}}$$

Answer

$$a + {{{b^2}} \over {c}}$$

Answer

$$a + {{{b^2}} \over {4c}}$$

Answer

$$a + {{{b^2}} \over {3c}}$$

Answer

$$a + {{{b^2}} \over {2c}}$$

Motion in 1D

The position of a particle as a function of time t, is given byx(t) = at + bt2 – ct3 where a, b and c are constants. When the particle attains zero acceleration, then its velocity will be :

Choose the correct answer: