Question 57: Motion in 1D

Question

The velocity of a particle is v = v0 + gt + Ft2. Its position is x = 0 at t = 0; then its displacement after time (t = 1) is :

Accepted Answer

The velocity of a particle is given by $ v = v_0 + gt + Ft^2 $. Its position is $ x = 0 $ at $ t = 0 $. To find its displacement after time $ t = 1 $, follow these steps:
Given:
$ v = v_0 + gt + Ft^2 $
We know that:
$ \frac{dx}{dt} = v_0 + gt + Ft^2 $
To find the displacement, integrate both sides with respect to $ t $:
$ \int_{x = 0}^{x} dx = \int_{t = 0}^{t = 1} (v_0 + gt + Ft^2) \, dt $
This simplifies to:
$ x = \left[ v_0 t + \frac{gt^2}{2} + \frac{Ft^3}{3} \right]_{t = 0}^{t = 1} $
Evaluating the integral from $ t = 0 $ to $ t = 1 $:
$ x = v_0 + \frac{g}{2} + \frac{F}{3} $
Therefore, the displacement after time $ t = 1 $ is:
$ x = v_0 + \frac{g}{2} + \frac{F}{3} $

Answer

v0 + g + f

Answer

v0 + $${g \over 2}$$ + $${F \over 3}$$

Answer

v0 + 2g + 3F

Answer

v0 + $${g \over 2}$$ + F

Motion in 1D

The velocity of a particle is v = v0 + gt + Ft2. Its position is x = 0 at t = 0; then its displacement after time (t = 1) is :

Choose the correct answer: