Question 20: Motion in 1D

Question

A particle is moving in a straight line. The variation of position '$$x$$' as a function of time '$$t$$' is given as $$x=\left(t^3-6 t^2+20 t+15\right) m$$. The velocity of the body when its acceleration becomes zero is :

Accepted Answer

The position equation is given by:
$$x = t^3 - 6t^2 + 20t + 15$$
First, compute the velocity $$v$$ by differentiating the position function with respect to time:
$$ \frac{d x}{d t} = v = 3t^2 - 12t + 20 $$
Next, compute the acceleration $$a$$ by differentiating the velocity function with respect to time:
$$ \frac{d v}{d t} = a = 6t - 12 $$
We need to find the time $$t$$ when the acceleration is zero:
$$ 6t - 12 = 0 $$
$$ t = 2 \, \mathrm{sec} $$
Now, find the velocity at $$t = 2 \, \mathrm{sec}$$:
$$ v = 3(2)^2 - 12(2) + 20 $$
$$ v = 8 \, \mathrm{m/s} $$

Answer

6 m/s

Answer

10 m/s

Answer

8 m/s

Answer

4 m/s

Motion in 1D

A particle is moving in a straight line. The variation of position '$x' as a function of time 't' is given as x=\left(t^3-6 t^2+20 t+15\right) m$. The velocity of the body when its acceleration becomes zero is :

Choose the correct answer: