Question 88: Motion in 1D

Question

The velocity of a particle is v = v0 + gt + ft2. If its position is x = 0 at t = 0, then its displacement after unit time (t = 1) is

Accepted Answer

Given that, v = v
0
+ gt + ft
2
We know that, $$v = {{dx} \over {dt}} $$
$$\Rightarrow dx = v\,dt$$
Integrating, $$\int\limits_0^x {dx}  = \int\limits_0^t {v\,dt} $$
or $$\,\,\,\,\,x = \int\limits_0^t {\left( {{v_0} + gt + f{t^2}} \right)} dt$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left[ {{v_0}t + {{g{t^2}} \over 2} + {{f{t^3}} \over 3}} \right]_0^t$$
or $$\,\,\,\,\,x = {v_0}t + {{g{t^2}} \over 2} + {{f{t^3}} \over 3}$$
At $$t = 1,\,\,\,\,\,x = {v_0} + {g \over 2} + {f \over 3}.$$

Answer

v0 + g/2 + f

Answer

v0 + 2g + 3f

Answer

v0 + g/2 + f/3

Answer

v0 + g + f

Motion in 1D

The velocity of a particle is v = v0 + gt + ft2. If its position is x = 0 at t = 0, then its displacement after unit time (t = 1) is

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