Question 9: Motion in 1D

Question

A particle moves along the $x$-axis and has its displacement $x$ varying with time t according to the equation:
$$ x=\mathrm{c}_0\left(\mathrm{t}^2-2\right)+\mathrm{c}(\mathrm{t}-2)^2 $$
where $\mathrm{c}_0$ and c are constants of appropriate dimensions.
Then, which of the following statements is correct?

Accepted Answer

To determine the acceleration of the particle, we start by differentiating the displacement function with respect to time $ t $.
The displacement of the particle is given by:
$ x = c_0(t^2 - 2) + c(t - 2)^2 $
First, compute the velocity $ v $ by differentiating $ x $ with respect to $ t $:
$ v = \frac{dx}{dt} = \frac{d}{dt}[c_0(t^2 - 2) + c(t - 2)^2] $
This gives:
$ v = 2c_0 t + 2c(t - 2) $
Next, determine the acceleration $ a $ by differentiating the velocity function with respect to time $ t $:
$ a = \frac{dv}{dt} = \frac{d}{dt}[2c_0 t + 2c(t - 2)] $
Calculating this gives:
$ a = 2c_0 + 2c $
Hence, the acceleration of the particle is $ 2c_0 + 2c $.

Answer

the acceleration of the particle is $2\left(c+c_0\right)$

Answer

the acceleration of the particle is $2 c_0$

Answer

the acceleration of the particle is 2 c

Answer

the initial velocity of the particle is $4 c$

Motion in 1D

A particle moves along the x-axis and has its displacement x varying with time t according to the equation: $ x=\mathrm{c}_0\left(\mathrm{t}^2-2\right)+\mathrm{c}(\mathrm{t}-2)^2 where \mathrm{c}_0$ and c are constants of appropriate dimensions. Then, which of the following statements is correct?

Choose the correct answer: